(Solved): Sodium metal reacts with water to produce hydrogen gas according to the following equation: \[ 2 \m ...

Sodium metal reacts with water to produce hydrogen gas according to the following equation: \[ 2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) \] The product gas, \( \mathrm{H}_{2} \), is collected over water at a temperature of \( 25^{\circ} \mathrm{C} \) and a pressure of \( 757.0 \mathrm{~mm} \mathrm{Hg} \). If the wet \( \mathrm{H}_{2} \) gas formed occupies a volume of \( 6.30 \mathrm{~L} \), the number of grams of \( \mathrm{H}_{2} \) formed is \( 23.8 \mathrm{~mm} \mathrm{Hg} \) at \( 25^{\circ} \mathrm{C} \). \[ 2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g) \] The product gas, \( \mathrm{H}_{2} \), is collected over water at a temperature of \( 25^{\circ} \mathrm{C} \) and a pressure of \( 748.0 \mathrm{~mm} \mathrm{Hg} \). If the wet \( \mathrm{H}_{2} \) gas formed occupies a volume of \( 8.63 \mathrm{~L} \), the number of moles of Al reacted was water is \( 23.8 \mathrm{~mm} \mathrm{Hg} \) at \( 25^{\circ} \mathrm{C} \). \[ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \] The product gas, \( \mathrm{C}_{2} \mathrm{H}_{2} \), is collected over water at a temperature of \( 20^{\circ} \mathrm{C} \) and a pressure of \( 758.0 \mathrm{~mm} \mathrm{Hg} \). If the wet \( \mathrm{C}_{2} \mathrm{H}_{2} \) gas formed occupies a volume of \( 7.56 \mathrm{~L} \), the number of grams of \( \mathrm{C}_{2} \mathrm{H}_{2} \) formed is water is \( 17.5 \mathrm{~mm} \mathrm{Hg} \) at \( 20^{\circ} \mathrm{C} \).