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Molar Volume of Gas Calculations – Answers
Answer One:
a)
CaCO3(s)  CaO(s) + CO2(g)
b)
relative molecular mass of CaCO3 = 40.0 + 12.0 + (3  16.0) = 100
moles = mass in grams  relative molecular mass
moles of CaCO3 = 25.0  100 = 0.250 mol
from balanced chemical equation, 1 mol CaCO3 produces 1 mol CO2
 0.250 mol CaCO3 produces 1/1  0.250 = 0.250 mol CO2
volume of gas in dm3 = moles  24.0
volume of CO2 = 0.250  24.0 = 6.00 dm3 (3 s.f.)
Answer Two:
a)
MgCO3(s) + 2HNO3(aq)  Mg(NO3)2(aq) + CO2(g) + H2O(l)
b)
moles of gas = volume in dm 3  24.0
moles of CO2 = 36.0  24.0 = 1.50 mol
from balanced chemical equation, 1 mol of CO2 is produced by 1 mol of MgCO3
 1.50 mol of CO2 is produced by 1/1  1.50 = 1.50 mol of MgCO3
relative molecular mass of MgCO3 = 24.0 + 12.0 + (3  16.0) = 84.0
mass in grams = moles  relative molecular mass
mass of MgCO3 = 1.50  84.0 = 126 g (3 s.f.)
Answer Three:
a)
MnO2(s) + 4HCl(aq)  MnCl2(aq) + 2H2O(l) + Cl2(g)
b)
relative molecular mass of MnO2 = 55.0 + (2  16.0) = 87.0
moles = mass in grams  relative molecular mass
moles of MnO2 = 8.7  87.0 = 0.100 mol
from balanced chemical equation, 1 mol MnO2 produces 1 mol Cl2
 0.100 mol MnO2 produces 1/1  0.100 = 0.100 mol Cl2
volume of gas in dm3 = moles  24.0
volume of Cl2 = 0.100  24.0 = 2.40 dm3 (3 s.f.)
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Answer Four:
moles of gas = volume in dm 3  24.0
moles of CO2 = 6.0  24.0 = 0.250 mol
from balanced chemical equation, 1 mol of CO2 is produced by 1 mol of M2CO3
 0.250 mol of CO2 is produced by 1/1  0.250 = 0.250 mol of M2CO3
0.250 mol of M2CO3 weighs 34.5 g
relative molecular mass = mass in grams  moles
 relative molecular mass of M2CO3 = 34.5  0.250 = 138
2  relative atomic mass of M = 138 – relative molecular mass of CO32–
2  relative atomic mass of M = 138 – (12.0 + (3  16.0)) = 78.0
relative atomic mass of M = 78.0  2 = 39.0 (3 s.f.)
M is therefore the Group I metal potassium (symbol, K)
 Group I metal carbonate = K2CO3
Answer Five:
a)
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
b)
relative molar mass of C3H8 = (3  12.0) + (8  1.0) = 44.0
moles = mass in grams  relative molecular mass
moles of C3H8 = 44.0  44.0 = 1.00 mol
relative molar mass of O2 = 2  16.0 = 32.0
moles of O2 = 80.0  32 = 2.50 mol (3 s.f.)
from the balanced chemical equation, 1 mol of C3H8 reacts with 5 mol of O2
however, there are only 2.50 mol of O2 present,  O2 is the limiting reagent for the reaction
alternatively
from the balanced chemical equation, 2.5 mol of O2 reacts with 0.5 mol of C3H8
however, there is 1.0 mol of C3H8 present,  C3H8 is in excess and O2 is the limiting reagent for the
reaction
from the balanced chemical equation 5 mol O2 produces 3 mol CO2
 2.50 mol O2 produces 2.50  3/5 = 1.50 mol CO2
volume of gas in dm3 = moles  24.0
volume of CO2 = 1.50  24.0 = 36.0 dm3 (3 s.f.)
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Answer Six:
a)
CaC2(s) + 2H2O(l)  Ca(OH)2(s) + C2H2(g)
b)
relative molecular mass of CaC2 = 40.0 + (2  12.0) = 64.0
moles = mass in grams  relative molecular mass
moles of CaC2 = 8.0  64.0 = 0.125 mol
from balanced chemical equation, 1 mol CaC2 produces 1 mol C2H2
 0.125 mol CaC2 produces 1/1  0.125 = 0.125 mol C2H2
volume of gas in dm3 = moles  24.0
volume of C2H2 = 0.125  24.0 = 3.00 dm3 (3 s.f.)
Answer Seven:
moles of gas = volume in dm 3  24.0
moles of H2 = 3.0  24.0 = 0.125 mol
from balanced chemical equation, 1 mol of H2 is produced by 1 mol of M
 0.125 mol of H2 is produced by 1/1  0.125 = 0.125 mol of M
0.125 mol of M weighs 5.0 g
relative atomic mass = mass in grams  moles
 relative atomic mass of M = 5.0  0.125 = 40.0 (3 s.f.)
 M is the Group II metal calcium (symbol, Ca)
Answer Eight:
a)
CH4(g) + H2O(g)  CO(g) + 3H2(g)
b)
moles of gas = volume in dm 3  24.0
moles of H2 = 1440  24.0 = 60.0 mol
from balanced chemical equation, 3 mol of H2 is produced by 1 mol of CH4
 moles of CH4 = 60.0  1/3 = 20.0 mol
volume of gas in dm3 = moles  24.0
volume of CH4 = 20.0  24.0 = 480 dm3 (3 s.f.)
Answer Nine:
a)
N2(g) + 3H2(g)  2NH3(g)
b)
moles of gas = volume in dm 3  24.0
moles of NH3 = 288  24.0 = 12.0 mol
from balanced chemical equation, 2 mol of NH3 are produced by 3 mol of H2
 12.0 mol of NH3 are produced by 12.0  3/2 = 18.0 mol of H2
volume of gas in dm3 = moles  24.0
volume of H2 = 18.0  24.0 = 432 dm3 (3 s.f.)
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Answer Ten:
a)
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
b)
moles of gas = volume in dm 3  24.0
moles of CO2 = 1.50  24.0 = 0.0625 mol
from balanced chemical equation, 1 mol of CO2 is produced by 1 mol of CaCO3
 0.0625 mol of CO2 is produced by 1/1  0.0625 = 0.0625 mol of CaCO3
relative molar mass of CaCO3 = 40.0 + 12.0 + (3  12.0) = 100
mass in grams = moles  relative molecular mass
mass of CaCO3 = 0.0625  100.0 = 6.25 g
% CaCO3 in rock = (mass of CaCO3  mass of rock)  100
% CaCO3 in rock = (6.25  8.00)  100 = 78.125 % (78.1 % to 3 s.f.)
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