A potential fuel for rockets is a combination of $\mathrm{B}_{5} \mathrm{H}_{9}\( and \)\mathrm{O}_{2}$ The two react according to the following balanced equation: $$ 2 \mathrm{B}_{5} \mathrm{H}_{9}(l)+12 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{B}_{2} \mathrm{O}_{3}(s)+9 \mathrm{H}_{2} \mathrm{O}(g) $$ If one tank in a rocket holds 126 \(\mathrm{g} \mathrm{B}_{5} \mathrm{H}_{9}\) and another tank holds \(192 \mathrm{g} \mathrm{O}_{2},\) what mass of water can be produced when the entire contents of each tank react together?

We have the following information: 1. Balanced chemical equation: $$ 2 \mathrm{B}_{5}\mathrm{H}_{9}(l)+12 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{B}_{2}\mathrm{O}_{3}(s)+9 \mathrm{H}_{2}\mathrm{O}(g) $$ 2. Mass of \(\mathrm{B}_{5}\mathrm{H}_{9}\) = \(126\,\text{g}\) 3. Mass of \(\mathrm{O}_{2}\) = \(192\,\text{g}\)

We need to convert the given masses of \(\mathrm{B}_{5}\mathrm{H}_{9}\) and \(\mathrm{O}_{2}\) into moles using their respective molar masses. Molar mass of \(\mathrm{B}_{5}\mathrm{H}_{9}\): \((5 \times 10.81) + (9 \times 1.008) = 54.05 + 9.072 = 63.122\,\text{g/mol}\) Molar mass of \(\mathrm{O}_{2}\): \(2 \times 16 = 32\,\text{g/mol}\) Moles of \(\mathrm{B}_{5}\mathrm{H}_{9} = \frac{126\,\text{g}}{63.122\,\text{g/mol}} = 1.996\,\text{mol}\) Moles of \(\mathrm{O}_{2} = \frac{192\,\text{g}}{32\,\text{g/mol}} = 6\,\text{mol}\)

According to the balanced chemical equation, \(2\,\text{mol}\) of \(\mathrm{B}_{5}\mathrm{H}_{9}\) react with \(12\,\text{mol}\) of \(\mathrm{O}_{2}\). We will now compare the reactants' given moles to their stoichiometric ratio. \(\frac{1.996\,\text{mol}\,\mathrm{B}_{5}\mathrm{H}_{9}}{2} = 0.998\) \(\frac{6\,\text{mol}\,\mathrm{O}_{2}}{12} = 0.5\) The value with the smaller result (0.5) indicates that \(\mathrm{O}_{2}\) is the limiting reactant.

Now that we know \(\mathrm{O}_{2}\) is the limiting reactant, we can determine the mass of water produced using the stoichiometry from the balanced equation: $$ 12\,\text{mol}\, \mathrm{O}_{2} \longrightarrow 9\,\text{mol}\, \mathrm{H}_{2}\mathrm{O} $$ Moles of \(\mathrm{H}_{2}\mathrm{O}\) produced: $$ \frac{9\,\text{mol}\, \mathrm{H}_{2}\mathrm{O}}{12\,\text{mol}\, \mathrm{O}_{2}} \times 6\,\text{mol}\, \mathrm{O}_{2} = 4.5\,\text{mol}\, \mathrm{H}_{2}\mathrm{O} $$ Molar mass of \(\mathrm{H}_{2}\mathrm{O}\): \((2 \times 1.008) + 16 = 18.016\,\text{g/mol}\) Now, we can calculate the mass of water produced: $$ 4.5\,\text{mol}\, \mathrm{H}_{2}\mathrm{O} \times 18.016\,\frac{\text{g}}{\text{mol}} = 81.072\,\text{g} $$

Therefore, a mass of \(81.072\,\text{g}\) of water (\(\mathrm{H}_{2}\mathrm{O}\)) can be produced when the entire contents of each tank react together.