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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?

Respuesta :

superman1987

Answer:

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.