Questions: Calcium hydride (CaH2) reacts with water to form hydrogen gas. CaH2(s) + 2 H2O(l) - Ca(OH)2(aq) + 2 H2(g) Determine the number of 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 37.6 g 0.894 g 151 g 43.4 g

Solution Steps

The reaction given is: \[ \text{CaH}_2(s) + 2 \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + 2 \text{H}_2(g) \] We need to determine the mass of \(\text{CaH}_2\) required to produce 45.0 L of \(\text{H}_2\) gas at 0.995 atm and \(32^\circ \text{C}\).

The ideal gas law is given by: \[ PV = nRT \] where:

• \(P = 0.995 \, \text{atm}\)
• \(V = 45.0 \, \text{L}\)
• \(R = 0.0821 \, \text{L atm/mol K}\)
• \(T = 32^\circ \text{C} = 305 \, \text{K}\)

We solve for \(n\), the number of moles of \(\text{H}_2\): \[ n = \frac{PV}{RT} = \frac{0.995 \times 45.0}{0.0821 \times 305} \]

\[ n = \frac{44.775}{25.0505} \approx 1.787 \, \text{mol} \]

From the balanced equation, 1 mole of \(\text{CaH}_2\) produces 2 moles of \(\text{H}_2\). Therefore, the moles of \(\text{CaH}_2\) needed are: \[ n_{\text{CaH}_2} = \frac{1.787}{2} \approx 0.8935 \, \text{mol} \]

The molar mass of \(\text{CaH}_2\) is: \[ \text{Ca} = 40.08 \, \text{g/mol}, \quad \text{H}_2 = 2 \times 1.008 \, \text{g/mol} = 2.016 \, \text{g/mol} \] \[ \text{Molar mass of } \text{CaH}_2 = 40.08 + 2.016 = 42.096 \, \text{g/mol} \]

The mass of \(\text{CaH}_2\) required is: \[ \text{mass} = n_{\text{CaH}_2} \times \text{molar mass} = 0.8935 \times 42.096 \approx 37.62 \, \text{g} \]

Final Answer