Calcium hydride reacts with water to form hydrogen gas according to the unbalanced equation below: CaH2 (s) + H2O (l) - -> Ca (OH) 2 (aq) + H2 (g) This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2 is desired. How many grams of calcium hydride are needed to generate 15.0 L of hydrogen gas at 25 degrees C and 825 torr of pressure?

28 grams CaH₂ (s) is required for production of 15L H₂ (g) at 25°C and 825Torr.

Explanation:

CaH₂ (s) + H₂O (l) = > Ca (OH) ₂ (s) + H₂ (g)

Using ideal gas law, PV = nRT

=> moles H₂ (g) = PV/RT = [ (825/760) Atm] (15L) / (0.08206L·Atm·mol⁻¹·K⁻¹) (298K) = 0.6659 mol H₂ (g)

From stoichiometry of given equation,

=> 0.6659 mol H₂ (g) requires 0.6659 mole CaH₂ (s)

Converting moles to grams, multiply by formula weight,

=> 0.6659 mole CaH₂ (s) = 0.6659 mole CaH₂ (s) x 42g/mole = 27.966 grams CaH₂ (s) ≅ 28 grams CaH₂ (s) (2 sig. figs.)