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 45.0 L of H2 gas at a pressure of 0.995 atm and a temperature of 32 °C?

37.67 g.

Explanation:

For the balanced reaction:

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

It is clear that 1 mol of CaH₂ (s) reacts with 2 mol of H₂O (l) to produce 1 mol of Ca (OH) ₂ (aq) and 2 mol of H₂ (g).

We need to calculate the no. of moles of generated 45.0 L of H2 gas at a pressure of 0.995 atm and a temperature of 32 °C:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 0.995 atm).

V is the volume of the gas in L (V = 45.0 L).

n is the no. of moles of the gas in mol (n = ? mol).

R is the general gas constant (R = 0.0821 L. atm/mol. K).

T is the temperature of the gas in K (T = 32.0° + 273 = 305.0 K).

∴ n = PV/RT = (0.0995 atm) (45.0 L) / (0.0821 L. atm/mol. K) (305.0 K) = 1.79 mol.

Using cross multiplication:

1 mol of CaH₂ (s) produces → 2 mol of H₂ (g), from stichiometry.

? mol of CaH₂ (s) produces → 1.79 mol of H₂ (g).

∴ no. of moles of CaH₂ (s) needed = (1 mol) (1.79 mol) / (2 mol) = 0.895 mol.

Now, we can get the mass of CaH₂ are needed:

∴ mass of CaH₂ = (no. of moles) (molar mass of CaH₂) = (0.895 mol) (42.094 g/mol) = 37.67 g.