Silane, \(\mathrm{SiH}_{4}\), is the silicon analogue of methane, \(\mathrm{CH}_{4}\). It is prepared industrially according to the following equations: $$ \begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g) \\ 4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l) \end{aligned} $$ a. If \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{~g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{~L}\) \(\mathrm{HCl}\) at \(10.0 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \%\) ?

First, we need to convert the volume and mass of \(\mathrm{HSiCl}_{3}\) to moles. The density of \(\mathrm{HSiCl}_{3}\) is given as \(1.34\, g/mL\). For 156 mL \(\mathrm{HSiCl}_{3}\), the mass would be: $$ \text{Mass(HSiCl}_{3}) = 156\, mL \times 1.34 \, g/mL = 209.04\, g $$ Now, find the molar mass of \(\mathrm{HSiCl}_{3}\). $$ \text{Molar Mass(HSiCl}_{3}) = 1\cdot M(\mathrm{H}) + 1\cdot M(\mathrm{Si}) + 3\cdot M(\mathrm{Cl}) = 1(1) + 1(28.1) + 3(35.5) = 97.6\, g/mol $$ We can now calculate the moles of \(\mathrm{HSiCl}_{3}\) as: $$ \text{Moles(HSiCl}_{3}) = \frac{\text{Mass(HSiCl}_{3})}{\text{Molar Mass(HSiCl}_{3})} =\frac{209.04\, g}{97.6\, g/mol} = 2.142\, mol $$ Next, we need to calculate the moles of \(\mathrm{HCl}\) for the given volume and conditions using the ideal gas law. $$ PV = nRT $$ We have the values: \(P = 10.0\,atm\), \(V = 15.0\,L\), \(T = 35^{\circ}\mathrm{C}\) which is equal to \(308.15\,K\) in Kelvin, and the gas constant, \(R = 0.0821\,\frac{L\cdot atm}{mol\cdot K}\). We can now find the moles of \(\mathrm{HCl}\): $$ n_\mathrm{HCl} = \frac{PV}{RT} =\frac{10.0\,atm \cdot 15.0\,L}{0.0821\frac {L\cdot atm}{mol\cdot K} \cdot 308.15\,K} = 6.264\,mol $$

From the balanced chemical equation, we have 1 mol of \(\mathrm{Si}\) reacting with \(3\) moles of \(\mathrm{HCl}\) to produce \(1\) mol of \(\mathrm{HSiCl}_{3}\). Thus, the reaction ratio of \(\mathrm{HCl}\) to \(\mathrm{HSiCl}_{3}\) is 3:1. The theoretical yield of \(\mathrm{HSiCl}_{3}\) would be: $$ \text{Theoretical moles(HSiCl}_{3}) = \frac{\text{Moles(HCl)}}{3} =\frac{6.264\, mol}{3} = 2.088\, mol $$

Now, we can determine the percent yield of \(\mathrm{HSiCl}_{3}\) as follows: $$ \% \text{Yield(HSiCl}_{3}) = \frac{\text{Moles(HSiCl}_{3})}{\text{Theoretical moles(HSiCl}_{3})}\times 100\% = \frac{2.142\, mol}{2.088\, mol}\times 100\% = 102.6\% $$

For part b, we are given that the percent yield of the reaction is 93.1%. We have the balanced equation which shows 4 moles of \(\mathrm{HSiCl}_{3}\) react to produce 1 mole of \(\mathrm{SiH}_{4}\): $$ \text{Actual moles(SiH4)} = \frac{\text{Actual moles(HSiCl}_{3})}{4} \times \% \text{Yield} = \frac{2.142\,mol}{4}\times(0.931) = 0.497\,mol $$

Now, we can calculate the volume of \(\mathrm{SiH}_{4}\) produced, using the ideal gas law and the given conditions (\(P = 10.0\, atm\) and \(T = 308.15\, K\)). $$ V_\mathrm{SiH4} =\frac{n_\mathrm{SiH4} RT}{P} = \frac{0.497\,mol \cdot 0.0821\frac{L\cdot atm}{mol\cdot K} \cdot 308.15\,K}{10.0\, atm} = 1.273\, L $$ So, when \(156\,mL\) of \(\mathrm{HSiCl}_{3}\) is heated, \(1.273\,L\) of \(\mathrm{SiH}_{4}\) will be obtained at \(10.0\,atm\) and \(35^{\circ} \mathrm{C}\), with a percent yield of \(93.1\%\).