Silane, SiH, , is the silicon analogue of methane, \(\mathrm{CH}_{4}\) . It is prepared industrially according to the following equations: $$\begin{array}{c}{\mathrm{Si}(s)+3 \mathrm{HCl}(g) \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g)} \\ {4 \mathrm{HSiCl}_{3}(l) \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l)}\end{array}$$ 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 \(156 \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is 93.1\(\% ?\)

= 5.947 moles of HCl#tag_title# Step 2: Calculate the theoretical yield of HSiCl3#tag_content# From the balanced equation, Si(s) + 3 HCl(g) -> HSiCl3(l) + H2(g) We can see that 1 mole of silicon reacts with 3 moles of HCl; thus, 5.947 moles of HCl theoretically react with 5.947/3 moles of silicon to produce 5.947/3 moles of HSiCl3. So, the theoretical yield of HSiCl3 is 5.947/3 moles = 1.982 moles. #tag_title# Step 3: Calculate the actual yield of HSiCl3 in moles#tag_content# We are given that 156 mL of HSiCl3 is produced with a density of 1.34 g/mL. We can convert this volume into mass and then into moles: mass = volume × density = 156 mL × 1.34 g/mL = 209.04 g of HSiCl3 Next, convert this mass to moles: moles = mass / molar mass = 209.04 g / (32.1 g/mol Si + 1.0 g/mol H + 35.5 g/mol Cl × 3) = 209.04 g / 169.6 g/mol = 1.232 moles #tag_title# Step 4: Calculate the percent yield of HSiCl3#tag_content# Percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100: percent yield = (actual yield / theoretical yield) × 100 = (1.232 moles / 1.982 moles) × 100 = 62.17 % #tag_title# Step 5: Calculate the moles of SiH4 produced using the second reaction#tag_content# We are given that the percent yield of the second reaction is 93.1%. We can use the actual yield of HSiCl3 obtained in Step 3 and the percent yield to find the moles of SiH4 produced: moles of SiH4 = moles of HSiCl3 × (percent yield / 100) = 1.232 moles × (93.1/100) = 1.147 moles SiH4 #tag_title# Step 6: Calculate the volume of SiH4 produced#tag_content# Now that we have the number of moles of SiH4, we can use the ideal gas law with the same temperature and pressure given for HCl to find the volume of SiH4 produced: V = nRT/P = (1.147 moles)(0.0821 L atm/mol K)(308.15 K) / 10.0 atm = 29.1 L So, the final answers are: a. The percent yield of HSiCl3 is \(62.17 \% \). b. The volume of SiH4 produced is \(29.1 L\) at 10.0 atm and 35°C.