Post on 17-Jan-2016
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Stoichiometry
Additional
Examples
HSTMr.Watson
Stoichiometry Flow Chart
mL soln A mL soln B (M/1000) \ Y moles B / (1000/M)
\ --------------- / 1/MM A \ X moles A / MM Bg A ---------------> mol A -----------------> mol B ---------------> g B / \ (P/1000RT) / (1000RT/P) \ / \ mL A gas mL B gas
where M => molarity of solutionMM => molar massP => pressure of gasR => gas constantT => temperature of gas
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, will react with 3.3
mol O2 to produce water?
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, will react with 3.3
mol O2 to produce water?
2 H2 + O2 -----> 2 H2O
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, will react with 3.3
mol O2 to produce water?
2 H2 + O2 -----> 2 H2O
(3.3 mol O2)
#g H2 = --------------------------------------------
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, will react with 3.3
mol O2 to produce water?
2 H2 + O2 -----> 2 H2O
(3.3 mol O2)
#g H2 = -------------------------------------------- (1 mol O2)
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, will react with 3.3
mol O2 to produce water?
2 H2 + O2 -----> 2 H2O
(3.3 mol O2) (2 mol H2)
#g H2 = -------------------------------------------- (1 mol O2)
stoichiometricfactor
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, will react with 3.3
mol O2 to produce water?
2 H2 + O2 -----> 2 H2O
(3.3 mol O2) (2 mol H2) (2.0 g H2)
#g H2 = -------------------------------------------- (1 mol O2) (1 mol H2)
the molar mass of H2
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, will react with 3.3
mol O2 to produce water?
2 H2 + O2 -----> 2 H2O
(3.3 mol O2) (2 mol H2) (2.0 g H2)
#g H2 = -------------------------------------------- (1 mol O2) (1 mol H2)
= 13 g H2
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
2 H2 + O2 -----> 2 H2O
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
2 H2 + O2 -----> 2 H2O
(5.40 g H2O)
#g H2 = --------------------
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
2 H2 + O2 -----> 2 H2O
(5.40 g H2O) (1 mol H2O)
#g H2 = ---------------------------------------------(18.0 g H2O)
molar mass of H2O
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
2 H2 + O2 -----> 2 H2O
(5.40 g H2O) (1 mol H2O) (2 mol H2)
#g H2 = --------------------------------------------- (18.0 g H2O)
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
2 H2 + O2 -----> 2 H2O
(5.40 g H2O) (1 mol H2O) (2 mol H2)
#g H2 = --------------------------------------------- (18.0 g H2O) (2 mol H2O)
stoichiometricfactor
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2)
#g H2 = ---------------------------------------------------------------
(18.0 g H2O) (2 mol H2O) (1 mol H2)
molar mass H2
HSTMr.Watson
EXAMPLEWhat mass of H2, in grams, must react with
excess O2 to produce 5.40 g H2O?
2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2)
#g H2 = ---------------------------------------------------------------
(18.0 g H2O) (2 mol H2O) (1 mol H2)
= 0.606 g H2
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?
KClO3 -----> O2 + ?
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?
KClO3 -----> O2 + ?
? => KCl
KClO3 -----> O2 + KCl
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?
KClO3 -----> O2 + ?
? => KCl
KClO3 -----> O2 + KCl
2 KClO3 -----> 3 O2 + 2 KCl
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 = -------------------------------------------------------------------------------
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 = -------------------------------------------------------------------------------
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 = -------------------------------------------------------------------------------
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass
stoichiometricfactor
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 = -------------------------------------------------------------------------------
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass molar mass
stoichiometricfactor
HSTMr.Watson
EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 = -------------------------------------------------------------------------------
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass molar mass
= 39.2 g O2 stoichiometricfactor
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber
process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber
process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
nitrogen + hydrogen -----> ammonia
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber
process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
nitrogen + hydrogen -----> ammonia
N2 + H2 -----> NH3
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber
process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
nitrogen + hydrogen -----> ammoniaN2 + H2 -----> NH3
N2 + 3 H2 -----> 2 NH3
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber
process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N2 + 3 H2 -----> 2 NH3
(3.0 kg NH3)
#mol H2 = -----------------
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber
process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N2 + 3 H2 -----> 2 NH3
(3.0 kg NH3) (1000 g)
#mol H2 = ------------------------------ (1 kg)
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber
process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N2 + 3 H2 -----> 2 NH3
(3.0 kg NH3) (1000 g) (1 mol NH3)
#mol H2 = -------------------------------------------- (1 kg) (17.0 g NH3)
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber process
by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N2 + 3 H2 -----> 2 NH3
(3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2)
#mol H2 = -------------------------------------------------------------
(1 kg) (17.0 g NH3) (2 mol NH3)
stoichiometricfactor
HSTMr.Watson
EXAMPLEAmmonia, NH3, is prepared by the Haber process
by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N2 + 3 H2 -----> 2 NH3
(3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2)
#mol H2 = -------------------------------------------------------------
(1 kg) (17.0 g NH3) (2 mol NH3)
= 2.6 X 102 mol H2
HSTMr.Watson
EXAMPLEThe first step in obtaining elemental zinc from
its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
HSTMr.Watson
EXAMPLEThe first step in obtaining elemental zinc from
its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
HSTMr.Watson
EXAMPLEThe first step in obtaining elemental zinc from
its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
ZnS + O2 -----> ZnO + SO2
HSTMr.Watson
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react
with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
ZnS + O2 -----> ZnO + SO2
2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2
HSTMr.Watson
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2)
#g O2 = ----------------------------------------------------------------
(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
stoichiometricfactor
HSTMr.Watson
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react
with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2)#g O2 = ----------------------------------------------------------------
(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
= 3.45 g O2
HSTMr.Watson
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2)
#gSO2 = ---------------------------------------------------------------
(97.44 g ZnS) (2 mol ZnS) (1 mol SO2)
HSTMr.Watson
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react
with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2)
#gSO2 = ---------------------------------------------------------------
(97.44 g ZnS) (2 mol ZnS) (1 mol SO2)
= 4.60 g SO2
HSTMr.Watson
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
#g O2 = 3.45 g O2
#gSO2 = 4.60 g SO2
HSTMr.Watson
EXAMPLEZinc and sulfur react to form zinc sulfide, a
substance used in phosphors that coat the inner surfaces of TV picture tubes. The equation for the reaction is Zn + S -----> ZnS In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and
allowed to react. Zn + S -----> ZnS
(a) Which is the limiting reactant? (b) How many grams of ZnS can be formed from this particular reaction mixture? (c) How many grams of which element will remain unreacted in this experiment?
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS)
#g ZnS = --------------------------------------------------------------
(65.38 g Zn) (1 mol Zn) (1 mol ZnS)
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS)
#g ZnS = --------------------------------------------------------------
(65.38 g Zn) (1 mol Zn) (1 mol ZnS)
= 17.9 g ZnS
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS
if use all 6.50 g S (6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS)
#g ZnS = -----------------------------------------------------------
(32.06 gS ) (1 mol S) (1 mol ZnS)
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS
if use all 6.50 g S (6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS)
#g ZnS = -----------------------------------------------------------
(32.06 gS ) (1 mol S) (1 mol ZnS)
= 19.8 g ZnS
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS
(a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = 17.9 g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.9 g ZnS can be produced.
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = 17.9 g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) How many grams of which element will remain unreacted in this experiment?
if use all 12.0 g Zn
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS?if use all 12.0 g Zn #g ZnS = 17.9 g ZnSif use all 6.50 g S #g ZnS = 19.8 g ZnSTherefore, Zn is the limiting reactant and 17.8 g
ZnS can be produced.
(c) if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S)
#g S = -------------------------------------------------------- (65.38 g Zn) (1 mol Zn) (1 mol S)
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = 17.9 g ZnSif use all 6.50 g S #g ZnS = 19.8 g ZnSTherefore, Zn is the limiting reactant and 17.8 g ZnS can
be produced.
(c) if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S)
#g S = -------------------------------------------------------- (65.38 g Zn) (1 mol Zn) (1 mol S)= 5.88 g S reacted
HSTMr.Watson
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS(a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = 17.9 g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn #g S = 5.88 g S reacted
Therefore, (6.50 - 5.88)g = 0.61 g S remain unreacted.
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
1(gaw)
%C = ------------ X 100
MM
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu= 119.377amu
1(gaw)%C = ------------ X 100
MM
1(12.011)%C = -------------- X 100 = 10.061% C
119.377
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
1(1.00797)
%H = ---------------- X 100 = 0.844359% H
119.377
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
3(35.453)
%Cl = -------------- X 100 = 89.095% Cl
119.377
HSTMr.Watson
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= 119.377amu%C = 10.061% C
%H = 0.844359% H
%Cl = 89.095% Cl