How many grams of calcium hydride are required to produce 1.14 L of hydrogen gas at 25.0°C and 0.975 atm pressure according to the chemical equation shown below? CaH₂(s)+ 2 H₂O(l)→ Ca(OH)₂(aq)+ 2 H₂(g) A) 0.956 g B) 1.91 g C) 3.82 g D) 11.4 g

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Quizplus · Accepted Answer

To solve this, we use the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin. First, convert temperature to Kelvin (25.0°C + 273.15 = 298.15 K) and use the given volume (1.14 L) and pressure (0.975 atm) to solve for n (moles of H2). Rearranging the ideal gas law gives n = PV/RT. Plugging in the values gives n = (0.975 atm * 1.14 L) / (0.0821 L·atm/mol·K * 298.15 K) ≈ 0.045 moles of H2. The stoichiometry of the reaction shows that 1 mole of CaH2 produces 2 moles of H2, so 0.045 moles of H2 would require half as many moles of CaH2, which is 0.0225 moles. The molar mass of CaH2 is approximately 42.09 g/mol, so the mass of CaH2 required is 0.0225 moles * 42.09 g/mol ≈ 0.956 g.

How Many Grams of Calcium Hydride Are Required to Produce