How many grams of CaH₂ are needed to generate 66.7 L of H₂ gas at a pressure of 0.995 atm and a temperature of 32°C? The molar mass of CaH₂ is 42.094 g/mol.

Question

Calcium hydride (CaH₂) reacts with water to form hydrogen gas: CaH₂(s) + 2H₂O(l) → Ca(OH)₂(aq) + 2H₂(g)

Answer 1

CaH₂(s) + 2H₂O(l) ==> Ca(OH)₂(aq) + 2H₂(g) ... balanced equation

First, convert 66.7 L H₂ gas into moles of H2 gas by using the ideal gas law:

PV = nRT

n = PV/RT = (0.995 atm)(66.7 L) / (0.0821 Latm/Kmol)(305K) note that Temp must be in Kelvin

n = 2.65 moles of H₂ gas

Using the stoichiometry of the balanced equation along with dimensional analysis, find the grams of CaH₂ needed:

2.65 mols H₂ x 1 mol CaH₂ / 2 mol H₂ x 42.094 g CaH₂/mol = 55.8 g CaH₂ needed