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Stoichiometry
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<= span style=3D'font-size:75%'>ØCalcula= te the amounts of reactants required or product formed in a nonchemical process.
<= span style=3D'font-size:75%'>ØInterpr= et balanced chemical equations in terms of interacting moles, representative particles, masse= s, and gas volume at STP.
<= span style=3D'font-size:75%'>ØConstru= ct mole ratios from balanced chemical equations and apply these ratios in mole-mole stoichiometric= calculations.
<= span style=3D'font-size:75%'>ØCalcula= te stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.
<= span style=3D'font-size:75%'>ØIdentif= y and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produce= 4d and the amount of excess reagent.
<= span style=3D'font-size:75%'>ØCalcula= te theoretical yield, or percent yield given appropriate information.
Ø
<= span style=3D'font-size:75%'>ØTEKS: 2= A, 2B, 2C, 2D, 2E, 3C, 3D, 7B, 11B, 11C
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Interpr= eting a Chemical Equation
<= span style=3D'font-size:75%'>ØStoichiomet= ry – calculations of quantities off the equation
<= span style=3D'font-size:75%'>ØFigure 9.4&= #13;
lN2<= /span>(g)   +   3H2<= /span>(g)   <= /span>→ 2NH3(g)
&Osla= sh;
------=_NextPart_01C5A277.C99A55B0 Content-Location: file:///C:/06899145/C-9Stoichiometrywebpage_files/slide0020.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Interpr= eting a Chemical Equation
<= span style=3D'font-size:75%'>ØStoichi= ometry – calculations of quantities off the equation
<= span style=3D'font-size:75%'>ØFigure 9.4         N2(g)   +   3H2<= /span>(g)   <= /span>→ 2NH3(g)
Ø     ▪ 1 molec N2  + 3= molec H2<= /span>  Ͱ= 0;   2 molec NH3
Ø   &n= bsp;
Ø   &n= bsp; ▪ 1( 6.02 x 1023) molec N2 +  3(6.02 x 1023) molec H2  ≠
Ø        = ;            &n= bsp;            = ;            &n= bsp;           2 (6.02 x23) molec NH3
Ø   &n= bsp; ▪ 1 mol N2 + 3 moles H2   ≠= ;   2 mol NH3
Ø   &n= bsp;
Ø   &n= bsp; ▪ 28.0 g N2  +  6.0 g H2 =3D  34.0 g NH3
Ø        = ;            &n= bsp;            = ;     34.0 g reactants  =3D<= span style=3D'mso-spacerun:yes'>  34.0 g products
Ø   &n= bsp;
Ø   &n= bsp; ▪ 22.4 L N2  +  3(22.4 L) H2  ≠ 2 (22.4 L) NH3
Ø        = ;            &n= bsp;            89.6 L of reactants  ≠   44.8 L products
= Ø
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Mole to= Mole Calculations
&O= slash;9 a)  Write the 6 mole ratios that can be derived from this equation.  9b)  How many moles of aluminum are neede= d to form 3.7 mol A2O3<= /span>.
Ø 4Al(s)   +   3O2(g)  = →   2Al2O3<= /span>(s)
l
l = ;
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Mole to= Mole Calculations
&O= slash;9 a)  Write the 6 mole ratios that can be derived from this equation.  9b)  How many moles of aluminum are neede= d to form 3.7 mol A2O3<= /span>.
Ø 4Al(s)   +   3O2(g)  = →   2Al2O3<= /span>(s)
l
l4 mol Al     ;    4 mol Al      ;    3 mol O2
l3 mol O2         2 mol Al2O3     2 mol Al= 2O= 3
l
l3 mol O2     ;    2 mol Al2O= 3    ;    2 mol Al2O= 3
l4 mol Al        = ;   4 mol Al        = ;   ;    3 mol O= 2
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Mole to= Mole Calculations
&O= slash;9 a)  Write the 6 mole ratios that can be derived from this equation.  9b)  How many moles of aluminum are neede= d to form 3.7 mol A2O3<= /span>.
Ø   mol        = ;            &n= bsp;          3.7 mol
Ø    4Al(s)   +   3O2(g)  = →   2Al2O3<= /span>(s)
&Osla= sh;
&Osla= sh;
Ø      3.7 mol A= l2<= /span>O3     x     4 mol Al        = ;  =3D     7.4 mol Al
Ø        = ;            &n= bsp;            = ;      2 mol Al2O3&= #13;
Ø     
l
l = ;
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Ø10a.  According to the equation in problem 9, how many moles of oxygen are required to react completely w= ith 14.8 mol Al?
Ø        14.= 8 mol     ? mol&= #13;
Ø     4Al(s)   +   3O2(g)  = →   2Al2O3<= /span>(s)
Ø
Ø      14.8 mol Al         3 mol O2    =3D  11.1 mol O2&= #13;
Ø        = ;            &n= bsp;            4 mol Al
Ø
Ø
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Mass to= Mass Calculations
Ø#11 Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2).  How many grams of acetylene are produced by adding water to 5.00g of CaC2<= /span>.
Ø CaC2(s)  = +  2H2O(l) → C2<= /span>H2(g) + Ca(OH)2(aq)
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Mass to= Mass Calculations
<= span style=3D'font-size:75%'>Ø#11 Ace= tylene gas (C2H2<= /span>) is produced by adding water to calcium carbide (CaC2).  H= ow many grams of acetylene are produced by adding water to 5.00g of CaC2.
Ø     gfm = CaC2<= /span> :      40.1  + 2(12.0) =3D 64.6 g/mol  
Ø     gfm = C2<= /span>H2  :      2(12.0) + 2(1.0)  =3D  26.0 g/mol
Ø       5.00 g        = ;            &n= bsp;       ? g
Ø       CaC2(s)  +  2H2O(l) → C2H2<= /span>(g) + Ca(OH)2(aq)
Ø
Ø   5.00 g CaC2<= /span>  x  1 mol CaC2  x  1 mol C2H2  x  26.0 g C2H2<= /span>  =3D=
Ø        = ;            &n= bsp;        64.6 g CaC2    1 mol CaC2<= /span>   = ;  1 mol C2H2&= #13;
= Ø
Ø        = ;      2.028081123 g  =3D  2.03 g C2H2&= #13;
Ø
Ø
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ØUse the same equation, determine how many moles of= CaC2 are needed to react completely wit= h 49.0 g H2O.
Ø CaC2(s)  = +  2H2O(l) → C2<= /span>H2(g) + Ca(OH)2(aq)
Ø
Ø
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ØUse the= same equation, determine how many moles of CaC2 are needed to react completely with 49.0 g H2O.
<= span style=3D'mso-special-format:nobullet;display:none;color:#99FF99;mso-color-= index: 6;font-family:Wingdings;font-size:80%'>Ø     gfm CaC2 :      40.1  + 2(12.0) =3D 64.6 g/mol <= span style=3D'font-size:88%'>  
<= span style=3D'mso-special-format:nobullet;display:none;color:#99FF99;mso-color-= index: 6;font-family:Wingdings;font-size:80%'>Ø     gfm = H2<= /span>O:         2(1.0)  + 16.0  =3D  18.0 g/mol
= Ø   
= Ø       ? mol         49.0 g        = ;       
= Ø     CaC<= /span>2<= /span>(s)  = +  2H2O(l) → C2<= /span>H2(g) + Ca(OH)2(aq)
Ø
= Ø     49.0= g H2<= /span>O  x = 1 mol H2<= /span>O  x = 1 mol CaC2  =3D
= Ø        = ;            &n= bsp;      18.0 g H2O    2 mol H2<= /span>O
Ø
= Ø        = ;            &n= bsp;          1.3611111 =3D 1.36 mol CaC2
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Ø#13  = How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KClO3) ?&#= 13;
Ø 2KClO3 = (s)  = →  2KCl(s)   +   3O2(g)
Ø
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Ø#13  = How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KCl= O3<= /span>) ?
Ø gfm KClO3<= /span> : 39= .1 + 35.5 + 3(16.0)  =3D  122.6 g/mol
Ø
Ø     6.54 g        = ;            &n= bsp;            = ;    ? molecules
Ø     2KCl= O3 = (s)  = →  2KCl(s)   +   3O2(g)
Ø
Ø6.54 g KClO3<= /span>  x  1 mol KClO3<= /span>  x  3 mol O2      x    6.02 x 1023=  mole= cules O2<= span style=3D'mso-spacerun:yes'>  =3D
Ø        = ;            &n= bsp;            = ;       18.0 g KClO3<= /span>   = ;  2 mol KClO3      1 mol O2&= #13;
Ø        = ;            &n= bsp;          
Ø        = ;            &n= bsp;            = ;            &n= bsp;      =3D 4.816982055 x 1022 molecules O2<= /span>  = ;
Ø        = ;            &n= bsp;            = ;            &n= bsp;      =3D 4.82 x 1022 molecules O2&= #13;
Ø
Ø
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Ø#28 The last step in the production of nitric acid= is the reaction of nitro= gen dioxide with water.  How many= grams of nitrogen dioxide must react with water to= produce 5.00 x 1022 mole= cules of nitrogen monoxide?
Ø 3NO2(g)   +   H2<= /span>O(l)   = →   2HNO3<= /span>(aq)   +   NO(g)<= /div>
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<= span style=3D'font-size:63%'>Ø#28 The= last step in the production of nitric acid is the reaction of nitrogen <= /span>dioxide with water.  How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 mole= cules of nitrogen monoxide?
= Ø gfm NO2 :  14.0 + 2(16.0)  =3D  44.0 g/mol
= Ø      
<= span style=3D'mso-special-format:nobullet;display:none;color:#99FF99;mso-color-= index: 6;font-family:Wingdings;font-size:80%'>Ø      ? g        = ;            &n= bsp;            = ;            &n= bsp;           5.00 x 1022 molec   = ;  
= Ø   3NO2<= /span>(g)   +   H2<= /span>O(l)   = →   2HNO3<= /span>(aq)   +   NO(g)
= Ø  
= Ø5.00 x 1022 molec NO  x  1 mol NO        = ;          x  3 mol NO2<= /span>  x  46.0 g NO2  = ;
= Ø        = ;            &n= bsp;            = ;     6.02 x 1023 molec NO     1 mol NO        1 mol NO2
Ø
= Ø        = ;            &n= bsp;            = ;       =3D 11.46179402   =3D  11.5 g NO2
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Ø#16.  Phosphorus and hydrogen can be combined to form phosphine (PH3).  H= ow many liters of phosphine are formed when 0.42L of hydrogen reacts with phosphorus?
Ø P4(s)   +   6H2<= /span>(g)   <= /span>→   4PH3(g)
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<= span style=3D'font-size:88%'>Ø#16.  Phosphorus and hydrogen can be co= mbined to form phosphine (PH<= span style=3D'font-size:58%;position:relative;top:.37em;mso-text-raise:-25%'>3<= /span>).  H= ow many liters of phosphine are fo= rmed when 0.42L of hydrogen reacts with phosphorus?
= Ø
Ø        = ;             0.42 L        = ;   ? L
Ø    P4<= /span>(s)   +   6H2<= /span>(g)   <= /span>→   4PH3(g) =
Ø   
Ø  0.42 L H2<= /span>  x   1 mol H2<= /span>  x   4 mol PH3<= /span>  x  22.4 L PH3
Ø        = ;            &n= bsp;   22.4 L H2      6 mol H2<= /span>        1 mol PH3
Ø
Ø        = ;            &n= bsp;            = ;            &n= bsp;       =3D 0.28 L PH3
Ø
Ø
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Ø#19  Isopropyl alcohol (C3H7<= /span>OH) burn in air according to this equation.&nb= sp; Calculate the moles of oxygen needed to react with 3.40 mol C3H7OH .<= span style=3D'mso-spacerun:yes'>  Find the moles of each product formed when 3.40 mol C3<= /span>H7OH re= acts with oxygen.
Ø 2 C3<= /span>H7OH(l)=   +  9O2(g)  →  6CO2(g)  = +  8H2O(g)
<= span style=3D'font-size:75%;visibility:hidden'>Ø
Ø
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Ø#19  Isopropyl alcohol (C3H7<= /span>OH) burn in air according to this equation.&nb= sp; Calculate the moles of oxygen needed to react with 3.40 mol C3H7OH .<= span style=3D'mso-spacerun:yes'>  Find the moles of each product formed when 3.40 mol C3<= /span>H7OH re= acts with oxygen.
Ø 3.40 mol         ? mol        = ;  ? mol     = ;     ? mol
Ø   2 C3H7OH(l)=   +  9O2(g)  →  6CO2(g)  = +  8H2O(g)
Ø
Ø    3.40 mol C3<= /span>H7OH  x   9 mol O2<= /span>        = ;     =3D  15.3 mol O= 2&= #13;
Ø        = ;            &n= bsp;            = ;      2 mol C3H7<= /span>OH
Ø    3.40 mol C= 3<= /span>H7OH  x    6 mol CO2<= /span>         =3D  10.2 mol CO2&= #13;
Ø        = ;            &n= bsp;            = ;       2 mol C3H7<= /span>OH
Ø    3.40 mol C= 3<= /span>H7OH  x    8 mol H2<= /span>O         =3D  13.6 mol H= 2<= /span>O
Ø        = ;            &n= bsp;            = ;       2 mol C3H7<= /span>OH
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Ø#21 The combustion of acetylene gas is represented= by the following equatio= n. a) How many grams of CO2 and H2O are produced when 52.0 g C2H2<= /span> burn= s. b) How many grams of oxyge= n are required to burn 52.0 g C2H2<= /span>.  Us= ing your answers from a and b to show = that this equation obeys the law of conservation of mass.
Ø C2<= /span>H2(g)  + 5O2(g)  = →  4CO= 2<= /span>(g)  = +2H2<= /span>O(g) 
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Ø#21 The combustion of acetylene gas is represented by the following = equation. a) How many grams of CO2<= /span> and = H2<= /span>O are produced when 52.0 g C2H2<= /span> burn= s. b) How many grams of oxygen are required to burn 52.0 g C2H2<= /span>.  Us= ing your answers from a and b to show that this equation obeys = the law of conservation of mass.
Ø  gfm CO2<= /span> : 12= + 2(16.0) =3D 44.0 g/mol ;  gfm= H2<= /span>O :  = 2(1.0) + 16.0 =3D 18.0 g/mol
Ø  gfm O2<= /span> :  2(16.0) =3D  32.0 g/mol  ;  gfm C2H2 : 2(12.0) +  = 2(1.0) =3D  26.0 g/mol
Ø
Ø 52.0 g        = ;  ? g        = ;       ? g        = ;  ? g
Ø    2C<= span style=3D'font-size:41%;position:relative;top:.37em;mso-text-raise:-25%'>2<= /span>H2(g)  + 5O2(g)  = →  4CO= 2<= /span>(g)  = +2H2<= /span>O(g)
Ø 
Ø   
Ø   52.0 g C2H2<= /span>  x  1 mol C2H2   x   4 mol CO2<= /span>   x   44.0g CO2<= /span> =3D = 176 g CO2&= #13;
Ø        = ;            &n= bsp;            = ;        26.0 g C2<= /span>H2       2 m= ol C2<= /span>H2      1 mol CO<= /span>2&= #13;
Ø
Ø    52.0 g C2H2<= /span>  x  1 mol C2H2   x   2 mol H2<= /span>O   x   18.0 g H2<= /span>O =3D 36.0 g H2O
Ø        = ;            &n= bsp;       26.0 g C2H2<= /span>      2 mol H2<= /span>O        1 mol H2O
= Ø
Ø   52.0 g C<= span style=3D'font-size:41%;position:relative;top:.37em;mso-text-raise:-25%'>2<= /span>H2  x  1 mol C2H2<= /span>   x   5 mol O2<= /span>       x   32.0 g O2 = =3D 160. g O2
Ø        = ;            &n= bsp;       26.0 g C2H2<= /span>       2 m= ol C2<= /span>H2       1 m= ol O2<= /span>
Ø
Ø   52.0 g + 160. g  =3D  176 g + 36.0 g  ;   212 g reactants =3D 212 g p= roduct
Ø     
Ø    
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Ø#22 a) How many liters of HF are needed to produce= 9.40 L H2<= /span> at S= TP?  b) How many molecules of H are pr= oduced by the reaction of = tin with 20.0L HF at STP?  c) How= many grams of SnF2 can = be made by reacting 7.42 x 1024 molecules of HF with tin?
Ø Sn(s)   +   2HF(g)   →  SnF2(s)   +   H2<= /span>(g)
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<= span style=3D'font-size:75%'>Ø#22 a) = How many liters of HF are needed to produce 9.40 L H2 at STP?  b) How many molecules of H are pr= oduced by the reaction of = tin with 20.0L HF at STP?  c) How= many grams of SnF2 can = be made by reacting 7.42 x 1024 molecules of HF with tin?
Ø    gfm SnF2:  118.7 + 2 (19.0) =3D 156.7 g/mol&= #13;
Ø        = ;            &n= bsp; ? L        = ;            &n= bsp;           9.40 L
Ø    Sn(s)   +   2HF(g)   →  SnF2(s)   +   H2<= /span>(g)
= Ø
Ø9.40 L H2  x  1 mol H2  x 2 mol HF x  22.4 L HF =3D  18.8 L HF
Ø        = ;            &n= bsp; 22.4 L H2    1 mol H2<= /span>      1 mol HF&= #13;
Ø
Ø20.0L HF  x 1 mol HF  x  1 mol H2  x  6.02 x 1023 molec H2 =3D  &n= bsp;  2.69 x 1024  =
Ø        = ;           22.4 L HF     2 mol HF     1 mol H2        = ;            &n= bsp;            = ;             molec H2&= #13;
= Ø
Ø7.42 x 1024 molec HF x  1 mol HF        = ;       x  1 mol SnF2<= /span>   x   156.7gSnF2<= /span> =3D<= span style=3D'mso-spacerun:yes'>        = ;       
Ø        = ;            &n= bsp;            = ;  6.02 x 1023 molec HF   2 mol HF        = ;    1 mol SnF2 &nb= sp;
Ø        = ;            &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;     =3D 966 g SnF2<= /span>
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Limiting Reagent
= ØLimiting Reagent – limits or determines the = amount of product. (It is t= he reactant that is used up first)
<= span style=3D'font-size:75%;visibility:hidden'>Ø
ØExcess Reag= ent – it is not completely used up in a chemical reaction
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= Ø#23 If 2.70 mol C2H4 is reacted with 6.30 mol O<= span style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2<= /span>,        = ;   a) identify the limiting reagent.  b) Calcula= te the moles   of water produced. c) Calculate the moles of exces= s reagent left.
Ø C2H4 (g)<= span style=3D'mso-spacerun:yes'>   +   3O2(g)  = →    2CO2(g)   +   2H2O(g)
Ø
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= Ø#23 If 2.70 mol C2H4 is reacted with 6.30 mol O<= span style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2<= /span>,        = ;   a) identify the limiting reagent.  b) Calcula= te the moles   of water produced. c) Calculate the moles of exces= s reagent left.
Ø  2.70 mol       6.3= 0 mol
Ø  C2H4 (g)<= span style=3D'mso-spacerun:yes'>   +   3O2(g)  = →    2CO2(g)   +   2H2O(g)
Ø  How to find limiting reagent - LR= 3;
Ø   2.70 mol C2<= /span>H4  =3D  2.70 mol C2H2<= /span>        = ;  Limiting Reagent=
Ø   1 coefficient        = ;            &n= bsp;            = ;            smallest # moles=
Ø   6.30 mol O2<= /span>   = ;  =3D   2.10 mol O<= /span>2&= #13;
Ø   3 coefficient &#= 13;
Ø
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= Ø#23 If 2.70 mol C2H4 is reacted with 6.30 mol O<= span style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2<= /span>,        = ;   a) identify the limiting reagent.  b) Calcula= te the moles   of water produced. c) Calculate the moles of exces= s reagent left.
Ø  ? mol excess  6. 30 mol        = ;            &n= bsp;    ? mol
Ø   C2H4 (g)<= span style=3D'mso-spacerun:yes'>   +   3O2(g)  = →    2CO2(g)   +   2H2O(g)
Ø
Ø  6.30 mol O2<= /span>  x  2 mol H2O  = =3D  4.20 mol H2<= /span>O
Ø        = ;            &n= bsp;       3 mol O2
Ø   6.30 mol O2<= /span>  x 1= mol C2<= /span>H4  =3D  2.10 mol C2H4&= #13;
Ø        = ;            &n= bsp;            = ;        3 mol O2&= #13;
Ø   2.70 mol C2<= /span>H4
Ø - 2.10 mol C<= /span>2<= /span>H4
Ø   0.60 mol C2<= /span>H4   (excess reagent)
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Ø#25 a) Identify the limiting reagent when 6.00 g HCl  reacts with 5.00 g Mg.  b) How many grams of hydrogen can be produced when 6.00 g HCl is added to 5.00 g= Mg?
Ø Mg(s)   +   2HCl(aq)   → MgCl<= /span>2<= /span>(aq)   +   H2<= /span>(g)
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Ø#25 a) Identify the limiting reagent when 6.00 g HCl  reacts with 5.00 g Mg.  = b) How many grams of hydrogen can be produced when 6.00 g HCl is added to 5.00 g Mg?
Ø 5.00 g        6.00 g        = ;            &n= bsp;            = ;  ? g        = ;            &n= bsp;           
Ø    Mg(s)   +   2HCl(aq)   → MgCl<= /span>2<= /span>(aq)   +   H2<= /span>(g)
Ø  
Ø 5.00 g Mg    x   1 mol Mg      =3D    0.206 mol Mg  =3D  0.206 mol Mg
Ø        = ;            &n= bsp;      24.3 g Mg    &= nbsp;      1 coefficient        = ;            &n= bsp;            = ;     LR
Ø    6.00 g HCl   x   1 mol HCl     =3D    0.164 mol HCl  =3D  0.082 mol HCl
Ø        = ;            &n= bsp;        36.5 g HCl    =        2 coefficient
Ø    Start calculation with LR
Ø    6.00 g HCl x  1 mol HCl x  1mol H2  x  2.0 g H2  =3D=   0.164g =3D 0.164 g H2 =
Ø        = ;            &n= bsp;    36.5 g HCl     2 mol HCl  1 mol H2&= #13;
Ø
Ø
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Percent= Yield
= l% Yield &nbs= p;  =3D     actual yield         x 100%        = ;            &n= bsp;       theoretic yield
l= 3;
l= 3;
lActual yield is usually given in the problem
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Ø27 When 84.8 g of iron (III) oxide reacts with an = excess of carbon monoxide,= 54.3 g of iron is produced.  What = is the percent yield of this reaction?
Ø Fe2<= /span>O3(s)   +   3CO(g)  →   2Fe(s)   +   3CO2(g)
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Ø27 When 84.8 g of iron (III) oxide reacts with an = excess of carbon monoxide, 54.3 g= of iron is produced.  What is the percent yield of this reacti= on?  Actual yield
Ø   84.8 g        = ;            &n= bsp;           ? g
Ø    Fe<= span style=3D'font-size:41%;position:relative;top:.37em;mso-text-raise:-25%'>2<= /span>O3(s)   +   3CO(g)  →   2Fe(s)   +   3CO2(g)
Ø        = ;            &n= bsp;            = ;             Theoretical yiel= d
Ø84.8 g Fe2O3<= /span>  x  1 mol Fe2O3   x   2 mol Fe      x    55.8 g Fe  =3D  59.3 g Fe   =
Ø        = ;            &n= bsp;     159.6 g Fe2O3<= /span>   = ;  1 mol Fe2O3<= /span>       1 m= ol Fe
Ø
Ø    % Yield  =3D     Actual Yield        x    100%= 3;
Ø        = ;            &n= bsp;  Theoretical Yield
Ø
Ø    % Yield  =3D  54.3 g    x<= span style=3D'mso-spacerun:yes'>   100%    =3D  91.6 %
Ø        = ;            &n= bsp; 59.3 g
Ø
Ø
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ØWhat is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum react with an excess of copper(II) sulfate? &nb= sp;
Ø 2Al(s)   +   3CuSO4<= /span>(aq)   = →  Al2<= /span>(SO4)3<= /span>(aq)   +   3Cu(s)=
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ØWhat is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum react with an excess of copper(II) sulfate? &nb= sp;         = ;            &n= bsp;    Actual Yield
Ø    1.87 g        = ;            &n= bsp;            = ;            &n= bsp;            = ;       ? g
Ø   2Al(s)   +   3CuSO4<= /span>(aq)  →   Al2(SO4)3<= /span>(aq)   +   3Cu(s)
Ø
Ø    1.87 g Al  x  1mol Al  x  3 mol Cu  x  63.5 g Cu   =3D   6.60 g Cu
Ø        = ;            &n= bsp;   27.0 g Al    2 mol Al       1 mol Cu
Ø        = ;            &n= bsp;            = ;            &n= bsp;          Theoretical Yiel= d
Ø   % Yield   =3D    Actual Yield       x   100%     
Ø        = ;            &n= bsp; Theoretical Yield
Ø   % Yield   =3D   4.65 g Cu    x  100%   =3D  70.5 %
Ø        = ;            &n= bsp;   6.60 g Cu
Ø 
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b3JkLXdyYXA6MTsNCgltc28tdmVydGljYWwtYWxpZ24tc3BlY2lhbDpiYXNlbGluZTsNCgltc28t YW5zaS1sYW5ndWFnZTpFTi1VUzt9DQphOmxpbmsNCgl7Y29sb3I6Izk5RkY5OSAhaW1wb3J0YW50 O30NCmE6YWN0aXZlDQoJe2NvbG9yOiMwMDg4RTQgIWltcG9ydGFudDt9DQphOnZpc2l0ZWQNCgl7 Y29sb3I6I0FGRTFGRiAhaW1wb3J0YW50O30NCn== ------=_NextPart_01C5A277.C99A55B0 Content-Location: file:///C:/06899145/C-9Stoichiometrywebpage_files/script.js Content-Transfer-Encoding: quoted-printable Content-Type: application/javascript; charset="us-ascii" function LoadSld() { var sld=3DGetObj("SlideObj") if( !g_supportsPPTHTML ) { =09 sld.style.visibility=3D"visible" return } if( MakeNotesVis() ) return runAnimations =3D _InitAnimations(); =09 if( IsWin("PPTSld") ) parent.SldUpdated(GetSldId()) g_origSz=3DparseInt(SlideObj.style.fontSize) g_origH=3Dsld.style.posHeight g_origW=3Dsld.style.posWidth g_scaleHyperlinks=3D(document.all.tags("AREA").length>0) if( g_scaleHyperlinks ) InitHLinkArray() if( g_scaleInFrame||(IsWin("PPTSld") && parent.IsFullScrMode() ) ) document.body.scroll=3D"no" _RSW() if( IsWin("PPTSld") && parent.IsFullScrMode() ) FullScrInit(); =09 MakeSldVis(); ChkAutoAdv() if( runAnimations ) { if( document.all("NSPlay") ) document.all("NSPlay").autoStart =3D false; if( sld.filters && sld.filters.revealtrans ) setTimeout( "document.body.start()", sld.filters.revealtrans.duration * = 1000 ); else document.body.start(); } } function MakeSldVis()=20 { var fTrans=3Dg_showAnimation && SldHasTrans() if( fTrans )=09 { if( g_bgSound ) { idx=3Dg_bgSound.indexOf(","); pptSound.src=3Dg_bgSound.substr( 0, idx ); pptSound.loop=3D -(parseInt(g_bgSound.substr(idx+1))); } SlideObj.filters.revealtrans.Apply()=09 } SlideObj.style.visibility=3D"visible" if( fTrans ) SlideObj.filters.revealtrans.Play() } function MakeNotesVis()=20 { if( !IsNts() ) return false=20 SlideObj.style.display=3D"none" nObj =3D document.all.item("NotesObj") parent.SetHasNts(0) if( nObj ) {=20 nObj.style.display=3D"" parent.SetHasNts(1) } return 1 } function ChkAutoAdv() { if(SldHasTrans()) SlideObj.onfilterchange=3DAutoAdv else AutoAdv() } function AutoAdv() { if(!IsWin("PPTSld") || !gUseSldTimings )return var sld=3DGetCurSld() if( (sld.mAdvDelay>0) && !parent.IsFramesMode() ) setTimeout("parent.GoToNextSld()",sld.mAdvDelay) } function GetObj(id) { if(g_supportsPPTHTML) return document.all(id); else return document.getElementById(id); } function SldHasTrans() { return SlideObj.style.filter !=3D ""; } function GetSldId() { return sId=3Dlocation.href.substring(location.href.la= stIndexOf('/')+1) } function HideMenu() { if( frames["PPTSld"] && PPTSld.document.all.item("ctx= tmenu") && PPTSld.ctxtmenu.style.display!=3D"none" ) { PPTSld.ctxtmenu.styl= e.display=3D'none'; return true } return false } function IsWin( name ) { return window.name =3D=3D name } function IsNts() { return IsWin("PPTNts") } function IsSldOrNts() { return( IsWin("PPTSld")||IsWin("PPTNts") ) } function SupportsPPTAnimation() { return( navigator.platform =3D=3D "Win32"= && navigator.appVersion.indexOf("Windows")>0 ) } function SupportsPPTHTML() { var appVer=3Dnavigator.appVersion, msie=3DappVer.indexOf("MSIE "), ver=3D0 if( msie >=3D 0 ) ver=3DparseFloat( appVer.substring( msie+5, appVer.indexOf(";",msie) ) ) else ver=3DparseInt(appVer) return( ver >=3D 4 && msie >=3D 0 ) } function _RSW() { if( !g_supportsPPTHTML || IsNts() || ( !g_scaleInFrame && (!IsWin("PPTSld") || !parent.IsFullScrMode()) ) ) return var padding=3D0; if( IsWin("PPTSld") && parent.IsFramesMode() ) padding=3D6 cltWidth=3Ddocument.body.clientWidth-padding cltHeight=3Ddocument.body.clientHeight-padding factor=3D(1.0*cltWidth)/g_origW if( cltHeight < g_origH*factor ) factor=3D(1.0*cltHeight)/g_origH newSize =3D g_origSz * factor if( newSize < 1 ) newSize=3D1 s=3DSlideObj.style s.fontSize=3DnewSize+"px" s.posWidth=3Dg_origW*factor s.posHeight=3Dg_origH*factor s.posLeft=3D(cltWidth-s.posWidth+padding)/2 s.posTop=3D(cltHeight-s.posHeight+padding)/2 if( g_scaleHyperlinks ) ScaleHyperlinks( factor ) } function _InitAnimations() { animRuntimeInstalled =3D ''+document.body.localTime !=3D 'undefined'; isFullScreen =3D (window.name =3D=3D "PPTSld") && !parent.IsFramesMode(); g_animUseRuntime =3D g_showAnimation && animRuntimeInstalled && !(isFullSc= reen && parent.IsSldVisited()); if( g_animUseRuntime ) { collSeq =3D document.all.tags("seq"); if( collSeq !=3D null ) { for(ii=3D0;ii numSlds ) gSldJumpIdx =3D numSlds; if ( gSldJumpIdx >=3D 0 ) { if ( gSldJumpIdx =3D=3D 0 ) gSldJumpIdx =3D 1; var jumpTo =3D parseInt(gSldJumpIdx); gSldJump =3D 0; gSldJumpIdx =3D ""; win.GoToSld( parent.GetSldList().mList[jumpTo-1].mSldHref ) } } } function _KDH() { if( event.keyCode =3D=3D 8 ) { event.returnValue =3D 0; parent.GoToPrevSld(); } } var g_HLinkArray =3D new Array(); =20 function IMapAreaObj( areaObj, coords ) { this.areaObj =3D areaObj; this.coords =3D coords; } function InitHLinkArray() { var appVer =3D navigator.appVersion; var msie =3D appVer.indexOf ( "MSIE " ) var ver =3D 0; if ( msie >=3D 0 ) ver =3D parseInt ( appVer.substring( msie+5 ) ); linkNum =3D 0; for (i=3D0; i 4 ) ) g_HLinkArray[linkNum++] =3D new IMapAreaObj( areaObj, ParseCoords( ar= eaObj.coords ) ); } } function ScaleHyperlinks( factor ) { =20 for ( ii=3D0; ii< g_HLinkArray.length; ii++) { coordsStr=3D""; imaObj =3D g_HLinkArray[ii]; for ( jj=3D0; jj < imaObj.coords.length-1; jj++ ) coordsStr +=3D (imaObj.coords[jj]*factor) + ","; coordsStr +=3D (imaObj.coords[jj]*factor); imaObj.areaObj.coords =3D coordsStr; } } function ParseCoords( coordsStr ) { var numArray =3D new Array(); var i =3D curPos =3D commaPos =3D 0; while ( ( commaPos =3D coordsStr.indexOf(",", curPos) ) !=3D -1 ) {=20 numArray[i++] =3D parseInt( coordsStr.substr( curPos, commaPos ) ); curPos =3D commaPos + 1; } numArray[i] =3D parseInt( coordsStr.substr( curPos ) ); return numArray; } function PlaySound( href, loop ){ href=3D unescape(href); if( window.event ) window.event.cancelBubble =3D true; pptSound.src =3D href; pptSound.loop =3D loop ? -1 : 0; } function DocumentOnClick() { if( IsNts() || parent.HideMenu() ) return; if( ( g_allowAdvOnClick && (window.name=3D=3D"PPTSld") && !parent.IsFrames= Mode() ) || (event && event.keyCode=3D=3D32) ) { =09 if( g_animUseRuntime && g_animMainSequence && g_animMainSequence.cangonex= t ) return; parent.GoToNextSld(); } } var g_supportsPPTHTML =3D SupportsPPTHTML(), g_scaleInFrame =3D 1, gId=3D""= , g_bgSound=3D"", g_scaleHyperlinks =3D false, g_allowAdvOnClick =3D 1, g_showInBrowser = =3D 0, gLoopCont =3D 0, gUseSldTimings =3D 1; var g_showAnimation =3D g_supportsPPTHTML && SupportsPPTAnimation() && ( (w= indow.name=3D=3D"PPTSld" && !parent.IsFramesMode()) || g_showInBrowser );va= r g_animManager =3D null; var g_animUseRuntime =3D false; var g_animItemsToHide, g_animInteractiveItems, g_animSlideTime; var g_animMainSequence =3D null; var ENDSHOW_MESG=3D"End of slide show, click to exit.", SCREEN_MODE=3D"Fram= es", gIsEndShow=3D0, NUM_VIS_SLDS=3D35, SCRIPT_HREF=3D"script.js", FULLSCR_= HREF=3D"fullscreen.htm"; var gCurSld =3D gPrevSld =3D 1, g_offset =3D 0, gNtsOpen =3D gHasNts =3D gO= tlTxtExp =3D 0, gHasNarration =3D 0, gOtlOpen =3D true window.gPPTHTML=3DSupportsPPTHTML() var gMainDoc=3Dnew Array(new hrefList("slide0001.htm",1,-1,1),new hrefList(= "slide0019.htm",1,-1,1),new hrefList("slide0002.htm",1,-1,1),new hrefList("= slide0020.htm",1,-1,1),new hrefList("slide0003.htm",1,-1,1),new hrefList("s= lide0021.htm",1,-1,1),new hrefList("slide0022.htm",1,-1,1),new hrefList("sl= ide0004.htm",1,-1,1),new hrefList("slide0005.htm",1,-1,1),new hrefList("sli= de0023.htm",1,-1,1),new hrefList("slide0024.htm",1,-1,1),new hrefList("slid= e0006.htm",1,-1,1),new hrefList("slide0007.htm",1,-1,1),new hrefList("slide= 0025.htm",1,-1,1),new hrefList("slide0008.htm",1,-1,1),new hrefList("slide0= 026.htm",1,-1,1),new hrefList("slide0009.htm",1,-1,1),new hrefList("slide00= 27.htm",1,-1,1),new hrefList("slide0010.htm",1,-1,1),new hrefList("slide002= 8.htm",1,-1,1),new hrefList("slide0011.htm",1,-1,1),new hrefList("slide0029= .htm",1,-1,1),new hrefList("slide0012.htm",1,-1,1),new hrefList("slide0031.= htm",1,-1,1),new hrefList("slide0013.htm",1,-1,1),new hrefList("slide0014.h= tm",1,-1,1),new hrefList("slide0032.htm",1,-1,1),new hrefList("slide0033.ht= m",1,-1,1),new hrefList("slide0034.htm",1,-1,1),new hrefList("slide0015.htm= ",1,-1,1),new hrefList("slide0016.htm",1,-1,1),new hrefList("slide0017.htm"= ,1,-1,1),new hrefList("slide0035.htm",1,-1,1),new hrefList("slide0018.htm",= 1,-1,1),new hrefList("slide0036.htm",1,-1,1)); /********************************************* Frameset functions These functions control slide navigation and state of the frameset. **********************************************/ function FullScrInit() { g_allowAdvOnClick =3D GetCurSld().mAdvOnClk document.body.style.backgroundColor=3D"black" document.oncontextmenu=3Dparent._CM; document.onkeydown =3D _KDH; document.ondragstart=3DCancel document.onselectstart=3DCancel self.focus() } function Redirect( frmId ) {=09 var str=3Ddocument.location.hash,idx=3Dstr.indexOf('#'), sId=3DGetSldId() if(idx>=3D0) str=3Dstr.substr(1); if( window.name !=3D frmId && ( sId !=3D str) ) { obj =3D GetObj("Main-File") window.location.href=3Dobj.href+"#"+sId return 1 } return 0 } var MHTMLPrefix =3D CalculateMHTMLPrefix();=20 function CalculateMHTMLPrefix() { if ( document.location.protocol =3D=3D 'mhtml:') {=20 href=3Dnew String(document.location.href)=20 Start=3Dhref.indexOf('!')+1=20 End=3Dhref.lastIndexOf('/')+1=20 if (End < Start)=20 return href.substring(0, Start)=20 else=20 return href.substring(0, End)=20 } return ''; } function GetTags(base,tag) { if(g_supportsPPTHTML) return base.all.tags(tag); else return base.getElementsByTagName(tag); } function UpdNtsPane(){ if(frames["PPTNts"]) PPTNts.location.replace( MHTMLP= refix+GetHrefObj( gCurSld ).mNtsHref ) } function UpdNavPane( sldIndex ){ if(gNavLoaded) PPTNav.UpdNav() } function UpdOtNavPane(){ if(gOtlNavLoaded) PPTOtlNav.UpdOtlNav() } function UpdOtlPane(){ if(gOtlLoaded) PPTOtl.UpdOtl() } function SetHasNts( fVal ) { if( gHasNts !=3D fVal ) { gHasNts=3DfVal UpdNavPane() } } function ToggleOtlText() { gOtlTxtExp=3D!gOtlTxtExp UpdOtlPane() } function ClearMedia() { // Clear any sounds playing before launching another browser window. Other= wise, // in fullscreen mode, you'll continue to hear the sound in the frames mod= e. if (PPTSld.pptSound) PPTSld.pptSound.loop =3D 0; } function FullScreen() {=20 if ( PPTSld.g_animUseRuntime ) PPTSld.document.body.pause(); ClearMedia(); var href =3D ( document.location.protocol =3D=3D 'mhtml:') ? FULLSCR_HREF = : FULLSCR_HREF+"#"+GetHrefObj(gCurSld).mSldHref; if(PPTNav.event.ctrlKey) { var w =3D (window.screen.availWidth * 1.0) / 2.0 var h =3D w * (PPTSld.g_origH * 1.0) / PPTSld.g_origW win =3D window.open( MHTMLPrefix+href,null,"toolbar=3D0,resizable=3D1,top= =3D0,left=3D0," + "width=3D"+ w + ",height=3D" + h ); if( win.document.body && PPTSld.g_animUseRuntime ) win.document.body.PPTSldFrameset=3Dwindow; } else { win =3D window.open( MHTMLPrefix+href,null,"fullscreen=3Dyes" ); if( win.document.body && PPTSld.g_animUseRuntime ) win.document.body.PPTSldFrameset=3Dwindow; } } function ToggleVNarration() { rObj=3DPPTSld.document.all("NSPlay") if( rObj && !PPTSld.g_animUseRuntime ) { if( (rObj.playState =3D=3D 1)||(rObj.playState =3D=3D 0) ) rObj.Play() else if( rObj.playState =3D=3D 2 ) rObj.Pause() else return; } else if( PPTSld.g_animUseRuntime ) { narObj =3D PPTSld.document.all("narrationID") if( narObj ) narObj.togglePause() } } function GetCurSldNum() { =20 obj=3DGetHrefObj(gCurSld) if( obj.mOrigVis =3D=3D 1 ) return obj.mSldIdx else =20 return gCurSld } function GetNumSlds() { =20 if( GetHrefObj(gCurSld).mOrigVis =3D=3D 1 ) return GetSldList().mNumVisSlds; else return GetSldList().mList.length } function GetSldNum( href ) { for(ii=3D0; ii 1 ) PopSldList(); else if( !IsFramesMode() ) { if( gLoopCont ) GoToFirst() else EndShow() } } function GoToPrevSld() { ii=3DgCurSld-1 if( ii > 0 ) { obj=3DGetHrefObj(ii) while ( obj && ( obj.mVis =3D=3D 0 ) && ( ii>0 ) ) obj=3DGetHrefObj(--ii) if( ii =3D=3D 0 ) ii=3D1 GoToSldNum(ii) } } function GoToFirst(){ GoToSld( GetHrefObj(1).mSldHref ) } function GoToLast() { ii=3DGetSldList().mList.length if( ii !=3D gCurSld ) GoToSld( GetHrefObj(ii).mSldHref ) } function GoToSldNum( num ) { if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue obj =3D GetHrefObj( num ) obj.mVis=3D1 gPrevSld=3DgCurSld gCurSld =3D num; PPTSld.location.replace(MHTMLPrefix+obj.mSldHref) if( IsFramesMode() ) { UpdNavPane(); UpdOtlPane(); UpdNtsPane() } } function GoToSld( href ) { if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue GetHrefObj( GetSldNum(href) ).mVis=3D1 PPTSld.location.replace(MHTMLPrefix+href) } function SldUpdated( id ) { if( id =3D=3D GetHrefObj(gCurSld).mSldHref ) return gPrevSld=3DgCurSld gCurSld=3DGetSldNum(id) if( IsFramesMode() ) { UpdNavPane(); UpdOtlPane(); UpdNtsPane() } } function PrevSldViewed(){ GoToSld( GetHrefObj(gPrevSld).mSldHref ) } function HasPrevSld() { return ( gIsEndShow || ( gCurSld !=3D 1 && GetHrefO= bj( gCurSld-1 ).mVis =3D=3D 1 )||( GetCurSldNum() > 1 ) ) } function HasNextSld() { return (GetCurSldNum() !=3D GetNumSlds()) } function CloseWindow() { if( HideMenu() ) return; =09 var event =3D PPTSld.event; if( !IsFramesMode() && event && (event.keyCode=3D=3D27 || event.keyCode=3D= =3D32 || event.type=3D=3D"click" ) ) window.close( self ); CatchNumKeys( self, event ); } function Unload() { gIsEndShow=3D0; } function SetupEndShow() { gIsEndShow=3D1; PPTSld.document.body.scroll=3D"no"; PPTSld.document.onkeypress=3DCloseWindow; PPTSld.document.onclick=3DCloseWindow; PPTSld.document.oncontextmenu=3D_CM; } function EndShow() { if( IsFramesMode() ) return if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue doc=3DPPTSld.document var dir =3D doc.body.dir if( dir !=3D "rtl" ) dir =3D "ltr"; doc.open() doc.writeln('


' + ENDSHOW_MESG + '

') doc.close() } function SetSldVisited(){ GetSldList().mList[gCurSld-1].mVisited=3Dtrue } function IsSldVisited(){ return GetSldList().mList[gCurSld-1].mVisited } function hrefList( sldHref, visible, advDelay, advClk ) { this.mSldHref=3D this.mNtsHref =3D sldHref this.mOrigVis=3D this.mVis =3D visible this.mVisited=3D false this.mAdvDelay=3D advDelay this.mAdvOnClk=3D advClk } function SldList(arr,curSld,fEnd) { this.mCurSld =3D curSld; this.mList =3D new Array(); var idx =3D 1; for(ii=3D0;ii 0) { PushSldList(sldList,fEnd); gCurSld =3D 1; } else if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue } function PushSldList(arr,fEnd) { var ii =3D gSldStack.length; gSldStack[ii] =3D new SldList(arr,gCurSld,fEnd); GoToSld( gSldStack[ii].mList[0].mSldHref ); } function PopSldList() { if (gSldStack[gSldStack.length-1].fEndShow) EndShow() else { gCurSld =3D gSldStack[gSldStack.length-1].mCurSld; gSldStack[gSldStack.length-1] =3D null; gSldStack.length--; var sldList =3D gSldStack[gSldStack.length-1]; GoToSld( sldList.mList[gCurSld - 1].mSldHref ); } } var custShowList=3Dnew Array(); /********************************************* Navigation button implementation There are 2 types of buttons: ImgBtn, TxtBtn implemented as function objects. They share a similiar interface so the event handlers can call SetActive, for example, on a button=20 object without needing to know exactly=20 what type of button it is. **********************************************/ //---------------------------------- function ImgBtn( oId,bId,w,action ) //---------------------------------- { var t=3Dthis t.Perform =3D _IBP t.SetActive =3D _IBSetA t.SetInactive=3D _IBSetI t.SetPressed =3D _IBSetP t.SetDisabled=3D _IBSetD t.Enabled =3D _IBSetE t.ChangeIcon =3D null t.UserAction =3D action t.ChgState =3D _IBUI t.mObjId =3D oId t.mBorderId=3D bId t.mWidth =3D w t.mIsOn =3D t.mCurState =3D 0 } function _IBSetA() { if( this.mIsOn ) { obj=3Dthis.ChgState( gHiliteClr,gShadowClr,2 ) obj.style.posTop=3D0 } } function _IBSetI() { if( this.mIsOn ) { obj=3Dthis.ChgState( gFaceClr,gFaceClr,1 ) obj.style.posTop=3D0=20 } } function _IBSetP() { if( this.mIsOn ) { obj=3Dthis.ChgState( gShadowClr,gHiliteClr,2 ) obj.style.posLeft+=3D1; obj.style.posTop+=3D1 } } function _IBSetD() { =20 obj=3Dthis.ChgState( gFaceClr,gFaceClr,0 ) obj.style.posTop=3D0=20 } function _IBSetE( state ) { var t=3Dthis GetObj( t.mBorderId ).style.visibility=3D"visible" if( state !=3D t.mIsOn ) { t.mIsOn=3Dstate if( state ) t.SetInactive() else t.SetDisabled() } } function _IBP() { var t=3Dthis if( t.mIsOn ) { if( t.UserAction !=3D null ) t.UserAction() if( t.ChangeIcon ) { obj=3DGetObj(t.mObjId) if( t.ChangeIcon() ) obj.style.posLeft=3Dobj.style.posLeft+(t.mCurState-4)*t.mWidth else obj.style.posLeft=3Dobj.style.posLeft+(t.mCurState-0)*t.mWidth } t.SetActive() } =20 } function _IBUI( clr1,clr2,nextState ) { var t=3Dthis SetBorder( GetObj( t.mBorderId ),clr1,clr2 ) obj=3DGetObj( t.mObjId ) obj.style.posLeft=3Dobj.style.posLeft+(t.mCurState-nextState)*t.mWidth-obj= .style.posTop t.mCurState=3DnextState return obj } //----------------------------------------- function TxtBtn( oId,oeId,action,chkState ) //----------------------------------------- { var t=3Dthis t.Perform =3D _TBP t.SetActive =3D _TBSetA t.SetInactive=3D _TBSetI t.SetPressed =3D _TBSetP t.SetDisabled=3D _TBSetD t.SetEnabled =3D _TBSetE t.GetState =3D chkState t.UserAction =3D action t.ChgState =3D _TBUI t.mObjId =3D oId t.m_elementsId=3D oeId t.mIsOn =3D 1 } function _TBSetA() { var t=3Dthis if( t.mIsOn && !t.GetState() ) t.ChgState( gHiliteClr,gShadowClr,0,0 ) } function _TBSetI() { var t=3Dthis if( t.mIsOn && !t.GetState() ) t.ChgState( gFaceClr,gFaceClr,0,0 ) } function _TBSetP() { if( this.mIsOn ) this.ChgState( gShadowClr,gHiliteClr,1,1 ) } function _TBSetD() { =20 this.ChgState( gFaceClr,gFaceClr,0,0 ) this.mIsOn =3D 0 } function _TBSetE() { var t=3Dthis if( !t.GetState() ) t.ChgState( gFaceClr,gFaceClr,0,0 ) else t.ChgState( gShadowClr,gHiliteClr,1,1 ) t.mIsOn =3D 1 } function _TBP() { var t=3Dthis if( t.mIsOn ) {=20 if( t.UserAction !=3D null ) t.UserAction() if( !t.GetState ) return if( t.GetState() ) t.SetPressed() else t.SetActive() } =20 } function _TBUI( clr1,clr2,lOffset,tOffset ) { SetBorder( GetObj( this.mObjId ),clr1,clr2 ) Offset( GetObj( this.m_elementsId ),lOffset,tOffset ) } function Offset( obj, top, left ){ obj.style.top=3Dtop; obj.style.left=3Dle= ft } function SetBorder( obj, upperLeft, lowerRight ) { s=3Dobj.style; s.borderStyle =3D "solid" s.borderWidth =3D 1=20 s.borderLeftColor =3D s.borderTopColor =3D upperLeft s.borderBottomColor=3D s.borderRightColor =3D lowerRight } function GetBtnObj(){ return gBtnArr[window.event.srcElement.id] } function BtnOnOver(){ b=3DGetBtnObj(); if( b !=3D null ) b.SetActive() } function BtnOnDown(){ b=3DGetBtnObj(); if( b !=3D null ) b.SetPressed() } function BtnOnOut(){ b=3DGetBtnObj(); if( b !=3D null ) b.SetInactive() } function BtnOnUp() { b=3DGetBtnObj() if( b !=3D null ) b.Perform() else Upd() } function GetNtsState(){ return parent.gNtsOpen } function GetOtlState(){ return parent.gOtlOpen } function GetOtlTxtState(){ return parent.gOtlTxtExp } function NtsBtnSetFlag( fVal ) { s=3Ddocument.all.item( this.m_flagId ).style s.display=3D"none" if( fVal ) s.display=3D"" else s.display=3D"none" } function _BSetA_Border(){ b =3D gBtnArr[this.mObjId]; if( b !=3D null ) b.S= etActive() } function _BSetI_Border(){ b =3D gBtnArr[this.mObjId]; if( b !=3D null ) b.S= etInactive() } function _BSetP_Border(){ b =3D gBtnArr[this.mObjId]; if( b !=3D null ) b.S= etPressed() } function _BSetA_BorderImg() {=20 b =3D gBtnArr[this.mBorderId]=20 if( b !=3D null && this.mIsOn && !b.GetState() ) { obj=3Dthis.ChgState( gHiliteClr,gShadowClr,2 ) obj.style.posTop=3D0 } } function _BSetI_BorderImg() {=20 b =3D gBtnArr[this.mBorderId] if( b !=3D null && this.mIsOn && !b.GetState() ) { obj=3Dthis.ChgState( gFaceClr,gFaceClr,1 ) obj.style.posTop=3D0 } } var gHiliteClr=3D"THREEDHIGHLIGHT",gShadowClr=3D"THREEDSHADOW",gFaceClr=3D"= THREEDFACE" var gBtnArr =3D new Array() gBtnArr["nb_otl"] =3D new TxtBtn( "nb_otl","nb_otlElem",parent.ToggleOtlPan= e,GetOtlState ) gBtnArr["nb_otlElem"] =3D new TxtBtn( "nb_otl","nb_otlElem",parent.ToggleOt= lPane,GetOtlState ) gBtnArr["nb_nts"] =3D new ImgBtn( "nb_nts","nb_ntsBorder",10,parent.ToggleN= tsPane ) gBtnArr["nb_nts"].SetActive =3D _BSetA_BorderImg; gBtnArr["nb_nts"].SetInactive =3D _BSetI_BorderImg; gBtnArr["nb_ntsBorder"] =3D new TxtBtn( "nb_ntsBorder","nb_ntsElem",parent.= ToggleNtsPane,GetNtsState ) gBtnArr["nb_ntsElem"] =3D new TxtBtn( "nb_ntsBorder","nb_ntsElem",parent.To= ggleNtsPane,GetNtsState ) gBtnArr["nb_prevBorder"] =3D gBtnArr["nb_prev"]=3D new ImgBtn( "nb_prev","n= b_prevBorder",30,parent.GoToPrevSld ) gBtnArr["nb_nextBorder"] =3D gBtnArr["nb_next"]=3D new ImgBtn( "nb_next","n= b_nextBorder",30,parent.GoToNextSld ) gBtnArr["nb_sldshw"]=3D new ImgBtn( "nb_sldshw","nb_sldshwBorder",18,parent= .FullScreen ) gBtnArr["nb_sldshwBorder"] =3D new TxtBtn( "nb_sldshw","nb_sldshwBorder",pa= rent.FullScreen,null ) gBtnArr["nb_sldshwBorder"].SetActive =3D _BSetA_Border; gBtnArr["nb_sldshwBorder"].SetInactive =3D _BSetI_Border; gBtnArr["nb_sldshwText"] =3D new TxtBtn( "nb_sldshw","nb_sldshwText",parent= .FullScreen,null ) gBtnArr["nb_sldshwText"].SetActive =3D _BSetA_Border; gBtnArr["nb_sldshwText"].SetInactive =3D _BSetI_Border; gBtnArr["nb_voice"] =3D gBtnArr["nb_voiceBorder"] =3D new ImgBtn( "nb_voice= ","nb_voiceBorder",18,parent.ToggleVNarration ) gBtnArr["nb_otlTxtBorder"] =3D gBtnArr["nb_otlTxt"]=3D new ImgBtn( "nb_otlT= xt","nb_otlTxtBorder",23,parent.ToggleOtlText ) gBtnArr["nb_ntsBorder"].m_flagId=3D "nb_nts" gBtnArr["nb_ntsBorder"].SetFlag =3D NtsBtnSetFlag gBtnArr["nb_otlTxt"].ChangeIcon=3D GetOtlTxtState /********************************************* Context menu implementation _CM() is the function that's hooked up to the oncontextmenu event. Once we're asked to show the menu, we first build it by creating DIVs on-the-fly. Then we position it=20 within the screen area so it doesn't get clipped. Creating the DIVs using createElement() means we don't have to write out any extra HTML into the slide HTML files. **********************************************/ var sNext=3D"Next",sPrev=3D"Previous",sEnd=3D"End Show",sFont=3D"Arial",sAr= row=3D"Arrow",sFreeform=3D"Freeform",sRect=3D"Rectangle",sOval=3D"Oval" function ShowMenu() { BuildMenu(); var doc=3DPPTSld.document.body,x=3DPPTSld.event.clientX+doc.scrollLeft,y= =3DPPTSld.event.clientY+doc.scrollTop m =3D PPTSld.document.all.item("ctxtmenu") m.style.pixelLeft=3Dx if( (x+m.scrollWidth > doc.clientWidth)&&(x-m.scrollWidth > 0) ) m.style.pixelLeft=3Dx-m.scrollWidth m.style.pixelTop=3Dy if( (y+m.scrollHeight > doc.clientHeight)&&(y-m.scrollHeight > 0) ) m.style.pixelTop=3Dy-m.scrollHeight m.style.display=3D"" } function _CM() { if( !parent.IsFullScrMode() ) return; if(!PPTSld.event.ctrlKey) { ShowMenu() return false } else HideMenu() } function BuildMenu() { if( PPTSld.document.all.item("ctxtmenu") ) return var mObj=3DCreateItem( PPTSld.document.body ) mObj.id=3D"ctxtmenu" mObj.style.visibility=3D"hidden" var s=3DmObj.style s.position=3D"absolute" s.cursor=3D"default" s.width=3D"120px" SetCMBorder(mObj,"menu","black") var iObj=3DCreateItem( mObj ) SetCMBorder( iObj, "threedhighlight","threedshadow" ) iObj.style.padding=3D2 CreateMenuItem( iObj,sNext,M_GoNextSld,M_True ) CreateMenuItem( iObj,sPrev,M_GoPrevSld,M_HasPrevSld ) =09 CreateSeparator( iObj ) CreateMenuItem( iObj,sEnd,M_End,M_True ) mObj.style.visibility=3D"visible" } function Cancel() { window.event.cancelBubble=3Dtrue; window.event.returnVa= lue=3Dfalse } function Highlight() { ChangeClr("activecaption","threedhighlight") } function Deselect() { ChangeClr("threedface","menutext") } function Perform() { e=3DPPTSld.event.srcElement if( e.type=3D=3D"menuitem" && e.IsActive() ) e.Action() else PPTSld.event.cancelBubble=3Dtrue } function ChangeClr( bg,clr ) { e=3DPPTSld.event.srcElement if( e.type=3D=3D"menuitem" && e.IsActive() ) { e.style.backgroundColor=3Dbg e.style.color=3Dclr } } function M_HasPrevSld() { return( parent.HasPrevSld() ) } function M_GoNextSld() { if( gIsEndShow ) M_End(); else GoToNextSld() } function M_GoPrevSld() { if( gIsEndShow ) { gIsEndShow=3D0; history.back();= PPTSld.event.cancelBubble=3Dtrue; } else GoToPrevSld() } function M_True() { return true } function M_End() { window.close( self ) } function CreateMenuItem( node,text,action,eval ) { var e=3DCreateItem( node ) e.type=3D"menuitem" e.Action=3Daction e.IsActive=3Deval e.innerHTML=3Dtext if( !e.IsActive() ) e.style.color=3D"threedshadow" e.onclick=3DPerform e.onmouseover=3DHighlight e.onmouseout=3DDeselect s=3De.style; 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charset="us-ascii" Slide 1 ------=_NextPart_01C5A277.C99A55B0 Content-Location: file:///C:/06899145/C-9Stoichiometrywebpage_files/outline.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii"
 No= tes
Slide Show
Outline
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