The reaction of calcium hydride with water can be used to prepare small quantities of hydrogen gas, as is done to fill weather-observation balloons. \(\mathrm{CaH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) $$ \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \text { (not balanced) } $$ (a) How many grams of \(\mathrm{H}_{2}(\mathrm{g})\) result from the reaction of \(127 \mathrm{g} \mathrm{CaH}_{2}\) with an excess of water? (b) How many grams of water are consumed in the reaction of \(56.2 \mathrm{g} \mathrm{CaH}_{2} ?\) (c) What mass of \(\mathrm{CaH}_{2}(\mathrm{s})\) must react with an excess of water to produce \(8.12 \times 10^{24}\) molecules of \(\mathrm{H}_{2} ?\)

The given chemical reaction is unbalanced. To balance it, compare the number of atoms on both sides and adjust the coefficients as necessary. The balanced chemical reaction is: \[ \mathrm{CaH}_{2}(\mathrm{s}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) + 2\mathrm{H}_{2}(\mathrm{g}) \]

(a) The molar mass of CaH2 is 42.094 g/mol. So, 127g of CaH2 is \(127 \, \mathrm{g} \times \left( \frac{1 \, \mathrm{mol}}{42.094 \, \mathrm{g}} \right) = 3.01 \, \mathrm{mol}\) Given that the mole ratio of CaH2 to H2 is 1:2, the moles of H2 produced is \(3.01 \, \mathrm{mol} \times 2 = 6.02 \, \mathrm{mol}\). The mass of H2 (with a molar mass of 2.02 g/mol) is then \(6.02 \, \mathrm{mol} \times \left( \frac{2.02 \, \mathrm{g}}{1 \, \mathrm{mol}} \right) = 12.16 \, \mathrm{g}\). (b) The molar mass of water (H2O) is 18.02 g/mol. For 56.2 g of CaH2, the moles is \(56.2 \, \mathrm{g} \times \left( \frac{1 \, \mathrm{mol}}{42.094 \, \mathrm{g}} \right) = 1.33 \, \mathrm{mol}\). The 1:2 mole ratio of CaH2 to H2O then gives \(2 \times 1.33 = 2.67 \, \mathrm{mol} \, H2O\). The mass of water consumed is thus \(2.67 \, \mathrm{mol} \times \left( \frac{18.02 \, \mathrm{g}}{1 \, \mathrm{mol}} \right) = 48.11 \, \mathrm{g}\). (c) The given number of H2 molecules can be converted to moles using Avogadro's number (6.02 x 10^23 molecules/mol); hence \(8.12 \times 10^{24} \, \mathrm{molecules} \times \left( \frac{1 \, \mathrm{mol}}{6.02 \times 10^{23} \, \mathrm{molecules}} \right) = 13.48 \, \mathrm{mol}\). Since the mole ratio of CaH2 to H2 is 1:2, the moles of CaH2 need is \(13.48/2 = 6.74 \, \mathrm{mol}\). Lastly, this is converted to mass using the molar mass of CaH2: \(6.74 \, \mathrm{mol} \times \left( \frac{42.094 \, \mathrm{g}}{1 \, \mathrm{mol}} \right) = 283.81 \, \mathrm{g} \)