Answer to Question #332558 in General Chemistry for micheal

Question

Calcium hydride, CaH2, reacts with water to form hydrogen gas:

CaH2 (s) + 2H2O (l) → Ca(OH)2 (aq) + 2H2 (g)

The 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 CaH2 are needed to generate 64.5 L of H2 gas if the pressure of H2 is 814 torr at 32 oC?

Answer 1

Solution:

Convert °C to K:

32°C + 273.15 = 305.15 K

Calculate the moles of hydrogen gas (H₂) using the ideal gas equation (PV = nRT):

n(H₂) = PV / RT

n(H₂) = (814 torr × 64.5 L) / (62.36 L torr mol⁻¹ K⁻¹ × 305.15 K) = 2.759 mol

Moles of H₂ = 2.759 mol

Balanced chemical equation:

CaH₂(s) + 2H₂O(l) → Ca(OH)₂(aq) + 2H₂(g)

According to stoichiometry:

1 mol of CaH₂ produces 2 mol of H₂

X mol of CaH₂ produces 2.759 mol of H₂

Thus,

Moles of CaH₂ = X = (2.759 mol H₂) × (1 mol CaH₂ / 2 mol H₂) = 1.3795 mol CaH₂

The molar mass of CaH₂ is 42.094 g/mol

Therefore,

Mass of CaH₂ = (1.3795 mol CaH₂) × (42.094 g CaH₂ / 1 mol CaH₂) = 58.07 g CaH₂ = 58.1 g CaH₂

Mass of CaH₂ = 58.1 g

Answer: 58.1 grams of CaH₂ are needed