Silicon nitride, \(\mathrm{Si}_{3} \mathrm{~N}_{4}\), is a ceramic material capable of withstanding high temperatures. It can be produced using the following unbalanced reaction: \(\mathrm{SiCl}_{4}+\mathrm{NH}_{3} \rightarrow \mathrm{Si}_{3} \mathrm{~N}_{4}+\mathrm{HCl}\) If \(64.2 \mathrm{~g}\) of \(\mathrm{SiCl}_{4}\) is reacted with \(20.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\), how many grams of \(\mathrm{Si}_{3} \mathrm{~N}_{4}\) are produced if the reaction has a \(96.0 \%\) yield?

After balancing the chemical equation, we have: 3\(\mathrm{SiCl}_{4}+12\mathrm{NH}_{3} \rightarrow \mathrm{Si}_{3} \mathrm{~N}_{4}+12\mathrm{HCl}\)

First, we need to find the molecular weights for the reactants: \(\mathrm{SiCl}_{4}: 28.09 + 4(35.45) = 169.9 \mathrm{~g/mol}\) \(\mathrm{NH}_{3}: 14.01 + 3(1.01) = 17.03 \mathrm{~g/mol}\) Next, we can convert the given grams of each reactant to moles: Moles of \(\mathrm{SiCl}_{4} = \frac{64.2 \mathrm{~g}}{169.9 \mathrm{~g/mol}} = 0.378 \mathrm{~mol}\) Moles of \(\mathrm{NH}_{3} = \frac{20.0 \mathrm{~g}}{17.03 \mathrm{~g/mol}} = 1.175 \mathrm{~mol}\)

We need to compare the mole ratios of the reactants to the balanced chemical equation: \(\frac {0.378 \mathrm{~mol} \; \mathrm{SiCl}_{4}}{3} \approx 0.126 \mathrm{~mol}\) \(\frac {1.175 \mathrm{~mol} \; \mathrm{NH}_{3}}{12} \approx 0.098 \mathrm{~mol}\) The smaller value of \(\frac{1.175}{12}\) indicates that \(\mathrm{NH}_{3}\) is the limiting reactant.

Now that we know the limiting reactant, \(\mathrm{NH}_{3}\), we can find the theoretical yield of \(\mathrm{Si}_{3} \mathrm{~N}_{4}\). Using the stoichiometric coefficients in the balanced equation, we can calculate the moles of \(\mathrm{Si}_{3} \mathrm{~N}_{4}\) produced: Moles of \(\mathrm{Si}_{3} \mathrm{~N}_{4} = \frac{1.175 \mathrm{~mol} \; \mathrm{NH}_{3}}{12} \times 1 = 0.098 \mathrm{~mol}\) Next, we can find the molecular weight of \(\mathrm{Si}_{3} \mathrm{~N}_{4}:\) \(\mathrm{Si}_{3} \mathrm{~N}_{4}: 3(28.09) + 4(14.01) = 140.3 \mathrm{~g/mol}\) Now, we can convert moles of \(\mathrm{Si}_{3} \mathrm{~N}_{4}\) to grams: Weight of \(\mathrm{Si}_{3} \mathrm{~N}_{4}= 0.098 \mathrm{~mol} \times 140.3 \mathrm{~g/mol} = 13.75 \mathrm{~g}\)

Now we need to adjust for the actual yield, which is \(96.0 \%.\) Actual mass of \(\mathrm{Si}_{3} \mathrm{~N}_{4}=0.960 \times 13.75 \mathrm{~g} = 13.2 \mathrm{~g}\)

The mass of \(\mathrm{Si}_{3} \mathrm{~N}_{4}\) produced with a \(96.0 \%\) yield is: \(13.2 \mathrm{~g}\)