Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$\mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g)$$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate $145 \mathrm{~L}\( of \)\mathrm{H}_{2}\( gas if the pressure of \)\mathrm{H}_{2}$ is \(110 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C} ?\)

Given the volume (V) of hydrogen gas = 145 L, pressure (P) = 110 kPa, and temperature (T) = 21°C. We need to convert the temperature to Kelvin (K) by adding 273.15 to the Celsius temperature: T(K) = 21 + 273.15 = 294.15 K. Now, we'll use the Ideal Gas Law: \(PV = nRT\) Where - P is the pressure in kPa - V is the volume in L - n is the number of moles - R is the ideal gas constant [8.314 J/(mol·K) or 0.08206 L·kPa/(mol·K)] - T is the temperature in K Rearrange the Ideal Gas Law to solve for the number of moles (n) of hydrogen gas (H₂): \(n = \frac{PV}{RT}\). Now, plug in the known values: P = 110 kPa, V = 145 L, R = 0.08206 L·kPa/(mol·K), and T = 294.15 K: \(n_{H2} = \frac{(110\: kPa)(145\: L)}{(0.08206\: L \cdot kPa/(mol \cdot K))(294.15\: K)}\)

According to the balanced chemical equation: 1 mole of CaH₂ (s) → 2 moles of H₂ (g) Now, use the stoichiometry to convert the moles of H₂ to moles of CaH₂: Let \(n_{CaH_2}\) be the number of moles of calcium hydride needed. \(\frac{n_{CaH_2}}{n_{H2}} = \frac{1}{2}\) \(n_{CaH_2} = \frac{1}{2} n_{H2}\) Now, calculate \(n_{CaH_2}\) using the value of \(n_{H2}\) obtained in the previous step: \(n_{CaH_2} = \frac{1}{2} (\frac{(110\: kPa)(145\: L)}{(0.08206\: L \cdot kPa/(mol \cdot K))(294.15\:K)})\)

To calculate the mass of calcium hydride needed, we need to know its molar mass, which is given by: Molar mass of CaH₂ = Molar mass of calcium (Ca) + 2 × Molar mass of hydrogen (H) Molar mass of CaH₂ = 40.08 g/mol (Ca) + 2 × 1.01 g/mol (H) = 42.10 g/mol Now, apply the formula mass = moles × molar mass: Mass of CaH₂ = \(n_{CaH_2} \times\) molar mass of CaH₂ \(m_{CaH_2} = \frac{1}{2} (\frac{(110\: kPa)(145\: L)}{(0.08206\: L \cdot kPa/(mol \cdot K))(294.15\:K)}) \times 42.10 \: g/mol\) Calculate the mass of calcium hydride needed to generate 145 L of hydrogen gas at the given pressure and temperature.