Question

Difficulty: HardHydrogen: Preparation, Properties, Isotopes, and Water Hardness

Calcium hydride (CaH2\text{CaH}_2) reacts vigorously with water to produce calcium hydroxide and hydrogen gas. What volume of dry hydrogen gas, in dm3\text{dm}^3, measured at standard temperature and pressure (s.t.p.), is liberated when 10.5 g10.5\text{ g} of pure calcium hydride reacts completely with excess water?

[Relative atomic masses: Ca=40\text{Ca} = 40, H=1\text{H} = 1; Molar volume of gas at s.t.p. = 22.4 dm3 mol122.4\text{ dm}^3\text{ mol}^{-1}]

Answer: 11.2 dm^3

Answer

The volume of dry hydrogen gas liberated at s.t.p. is 11.2 dm³.
Calcium hydride reacts with water according to the reaction CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. Given 10.5 g of CaH₂ (molar mass 42 g/mol), there are 0.25 moles of CaH₂. Based on the 1:2 stoichiometric ratio, 0.50 moles of H₂ gas are generated. Multiplying by the molar volume at s.t.p. (22.4 dm³/mol) yields 11.2 dm³.

Step-by-Step Solution

1
Write the balanced chemical equation for the reaction of calcium hydride with water
CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
Establishing the stoichiometric mole ratio between the reactant CaH₂ and the product H₂ gas.
2
Calculate the molar mass of CaH₂
42 g/mol
Molar mass is required to convert the given mass of CaH₂ into moles.
3
Determine the amount of CaH₂ in moles
0.25 mol
Moles = Mass / Molar mass = 10.5 g / 42 g/mol.
4
Calculate the moles of H₂ gas liberated using the 1:2 stoichiometric ratio
0.50 mol
1 mole of CaH₂ produces 2 moles of H₂ gas.
5
Calculate the volume of H₂ gas produced at standard temperature and pressure (s.t.p.)
11.2 dm³
Volume at s.t.p. = Moles × Molar volume at s.t.p. = 0.50 mol × 22.4 dm³/mol.

Key Concept

Laboratory and industrial preparation of hydrogen using metal hydrides and mole-volume stoichiometric calculations at s.t.p.
Estimated Time:2m 0s
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