For the reaction \(\mathrm{Ca}_{3} \mathrm{P}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{PH}_{3}\), (a) Balance the equation. (b) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) would you need to react with \(60.0 \mathrm{~g}\) of \(\mathrm{Ca}_{3} \mathrm{P}_{2} ?\) (c) How many grams of \(\mathrm{PH}_{3}\) could theoretically be produced from the reactant amounts calculated in part (b)?

In order to balance the chemical equation, adjust the coefficients of the reactants and products until there is the same amount of each element on both sides of the equation. Given chemical equation: \(\mathrm{Ca}_{3}\mathrm{P}_{2}+\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{PH}_{3}\) Balancing the equation: \(1\,\mathrm{Ca}_{3}\mathrm{P}_{2}+6\,\mathrm{H}_{2}\mathrm{O} \rightarrow 3\,\mathrm{Ca}(\mathrm{OH})_{2}+2\,\mathrm{PH}_{3}\)

To find out how many grams of \(\mathrm{H}_{2}\mathrm{O}\) are needed to react with \(60.0\,\mathrm{g}\) of \(\mathrm{Ca}_{3}\mathrm{P}_{2}\), first convert the mass of \(\mathrm{Ca}_{3}\mathrm{P}_{2}\) to moles using its molar mass. The molar mass of \(\mathrm{Ca}_{3}\mathrm{P}_{2}\) is \(3\times40.08\ \mathrm{g/mol}\) (for calcium) + \(2\times30.97\ \mathrm{g/mol}\) (for phosphorus) = \(310.38\,\mathrm{g/mol}\). Moles of \(\mathrm{Ca}_{3}\mathrm{P}_{2}\) = \(\frac{60.0\,\mathrm{g}}{310.38\,\mathrm{g/mol}} = 0.193\,\mathrm{mol}\)

Use the stoichiometric coefficients from the balanced chemical equation to determine the moles of \(\mathrm{H}_{2}\mathrm{O}\) needed to react with the moles of \(\mathrm{Ca}_{3}\mathrm{P}_{2}\) calculated in step 2. From the balanced equation, we see that \(1\,\text{mol}\,\mathrm{Ca}_{3}\mathrm{P}_{2}\) reacts with \(6\,\text{mol}\,\mathrm{H}_{2}\mathrm{O}\). Moles of \(\mathrm{H}_{2}\mathrm{O}\) needed = \(0.193\,\mathrm{mol}\,\mathrm{Ca}_{3}\mathrm{P}_{2}\times \frac{6\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}}{1\,\mathrm{mol}\,\mathrm{Ca}_{3}\mathrm{P}_{2}}=1.158\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}\)

To find how many grams of \(\mathrm{H}_{2}\mathrm{O}\) are needed, convert the moles of \(\mathrm{H}_{2}\mathrm{O}\) to grams using its molar mass. The molar mass of \(\mathrm{H}_{2}\mathrm{O}\) is \(2\times1.01\,\mathrm{g/mol}\) (for hydrogen) + \(1\times16.00\,\mathrm{g/mol}\) (for oxygen) = \(18.02\,\mathrm{g/mol}\). Grams of \(\mathrm{H}_{2}\mathrm{O}\) needed = \(1.158\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}\times \frac{18.02\,\mathrm{g}}{1\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}}= 20.84\,\mathrm{g}\,\mathrm{H}_{2}\mathrm{O}\)

Use the stoichiometric coefficients from the balanced chemical equation and the moles of \(\mathrm{Ca}_{3}\mathrm{P}_{2}\) to determine the moles of \(\mathrm{PH}_{3}\) that can be produced. Convert the moles of \(\mathrm{PH}_{3}\) to grams using its molar mass. From the balanced equation, we see that \(1\,\text{mol}\,\mathrm{Ca}_{3}\mathrm{P}_{2}\) reacts to produce \(2\,\text{mol}\,\mathrm{PH}_{3}\). Moles of \(\mathrm{PH}_{3}\) produced = \(0.193\,\mathrm{mol}\,\mathrm{Ca}_{3}\mathrm{P}_{2}\times \frac{2\,\mathrm{mol}\,\mathrm{PH}_{3}}{1\,\mathrm{mol}\,\mathrm{Ca}_{3}\mathrm{P}_{2}}=0.386\,\mathrm{mol}\,\mathrm{PH}_{3}\) The molar mass of \(\mathrm{PH}_{3}\) is \(1\times30.97\ \mathrm{g/mol}\) (for phosphorus) + \(3\times1.01\ \mathrm{g/mol}\) (for hydrogen) = \(34.00\,\mathrm{g/mol}\). Grams of \(\mathrm{PH}_{3}\) produced = \(0.386\,\mathrm{mol}\,\mathrm{PH}_{3}\times \frac{34.00\,\mathrm{g}}{1\,\mathrm{mol}\,\mathrm{PH}_{3}}= 13.12\,\mathrm{g}\,\mathrm{PH}_{3}\) To summarize, to react with \(60.0\,\mathrm{g}\) of \(\mathrm{Ca}_{3}\mathrm{P}_{2}\), the required amount of \(\mathrm{H}_{2}\mathrm{O}\) is \(20.84\,\mathrm{g}\), and the theoretical yield of \(\mathrm{PH}_{3}\) that can be produced is \(13.12\,\mathrm{g}\).