what volume in liters of h2 gas is produced at 1.25 atm and 58 c when 85 grams of CaH2 are reacted with 100 grams h2o?

Molar mass of CaH₂ = (40.1 + 1.0×2) g/mol = 42.1 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol

Initial no. of moles of CaH₂ = (85 g) / (42.1 g/mol) = 2.019 mol
Initial no. of moles of H₂O = (100 g) / (18.0 g/mol) = 5.556 mol

CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
Mole ratio CaH₂ : H₂O = 1 : 2

When CaH₂ completely reacts, no. of moles of H₂O = (2.019 mol) × 2 = 4.038 mol < 5.556 mol
Hence, H₂O is in excess, and CaH₂ completely reacts.

According to the above equation, mole ratio CaH₂ : H₂ = 1 : 2
No. of moles of CaH₂ reacted = 2.019 mol
No. of moles of H₂ formed = (2.019 mol) × 2 = 4.038 mol

For the H₂ formed :
No. of moles, n = 4.038 mol
Pressure, P = 1.25 atm
Temperature, T = (273 + 58) K = 331 K
Gas constant, R = 0.08206 atm L / (mol K)
Volume, V = ? L

PV = nRT
Volume of H₂ formed, V = nRT/P = 4.038 × 0.08206 × 331 / 1.25 = 87.7 L