MSJChem – Topic 1 – Molar volume of a gas
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Molar volume of a gas
One mole of a gas occupies 22.7 dm3 at STP (1.00 × 105 Pa and 273 K)
1) A sample of gas at STP contains 0.754 mol of Cl2. Calculate the following:
a) the volume occupied by the gas
b) the mass of Cl2 present
c) the number of Cl2 molecules in the sample of gas
d) the number of Cl atoms present in the sample
2) A sample of O2 gas at STP contains 3.01 × 1023 molecules. Calculate the
following:
a) the amount of O2 in mol
b) the mass of O2 present
c) the volume occupied by the gas
d) the number of oxygen atoms present in the sample
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MSJChem – Topic 1 – Molar volume of a gas
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3) A sample of N2 gas at STP has a mass of 25.0g. Calculate the following:
a) the amount of N2 in mol
b) the volume occupied by the gas
c) the number of nitrogen molecules present in the sample
4) A sample of gas at STP contains 5.72 mol of NH3. Calculate the following:
a) the volume occupied by the gas
b) the number of NH3 molecules present in the sample
c) the number of hydrogen atoms present in the sample
5) 3.54 g of magnesium is reacted with excess hydrochloric acid. Calculate the
volume of hydrogen gas produced at STP.
6) 139 g of calcium carbonate is reacted with excess hydrochloric acid.
Calculate the volume of carbon dioxide produced at STP.
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MSJChem – Topic 1 – Molar volume of a gas
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Answers:
1)
a) (0.754 × 22.7) = 17.1 dm3
b) (0.754 × 70.9) = 53.5g Cl2
c) (0.754 × 6.02 × 1023) = 4.54 × 1023 Cl2 molecules
d) (4.54 × 1023 × 2) = 9.08 × 1023 Cl atoms
2)
a) (3.01 × 1023 ÷ 6.02 × 1023) = 0.500 mol O2
b) (0.500 × 32.00) = 16.0g O2
c) (0.500 × 22.7) = 11.4 dm3
d) (3.01 × 1023 × 2) = 6.02 × 1023 oxygen atoms
3)
a) (25.0 ÷ 28.02) = 0.892 mol N2
b) (0.892 × 22.7) = 20.2 dm3 N2
c) (0.892 × 6.02 × 1023) = 5.37 × 1023 molecules N2
4)
a) (5.72 × 22.7) = 1.30 × 102 dm3
b) (5.72 × 6.02 × 1023) = 3.44 × 1024 molecules NH3
c) (3.44 × 1024 × 3) = 1.03 × 1025 atoms H
5) Balanced equation: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
(n) Mg = (3.54 ÷ 24.31) = 0.146 mol
Ratio of Mg to H2 is 1:1
(n) H2 = 0.146 mol
Volume of H2 at STP = (0.146 × 22.7) = 3.31 dm3
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6) Balanced equation: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
(n) CaCO3 = (139 ÷ 100.09) = 1.39 mol
Ratio CaCO3 to CO2 is 1:1
(n) CO2 = 1.39 mol
Volume of CO2 at STP = (1.39 × 22.7) = 31.6 dm3
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