Calcium hydride (cah2) reacts with water to form hydrogen gas: cah2 (s) + 2h2o (l) → ca (oh) 2 (aq) + 2h2 (g) how many grams of cah2 are needed to generate 58.0 l of h2 gas at a pressure of 0.811 atm and a temperature of 32°c?

Using the ideal gas law equation, we can find the number of H₂ moles produced.

PV = nRT

Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa

V - volume - 58.0 x 10⁻³ m³

R - universal gas constant - 8.314 Jmol⁻¹K⁻¹

T - temperature - 32 °C + 273 = 305 K

substituting these values in the equation,

82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K

n = 1.88 mol

The balanced equation for the reaction is as follows;

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

stoichiometry of CaH₂ to H₂ is 1:2

When 1.88 mol of H₂ is formed, number of CaH₂ moles reacted = 1.88/2 mol

therefore number of CaH₂ moles reacted = 0.94 mol

Mass of CaH₂ reacted - 0.94 mol x 42 g/mol = 39.48 g of CaH₂ are needed