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

Step by step solution

01

Converting the given volume of HSiCl3 to mass

Given that the density (\(d\)) of HSiCl3 equals 1.34 g/mL, the mass of HSiCl3 can be calculated by multiplying the density by the volume: \(volume \times density = mass\) or \(156 \, mL \times 1.34 \, g/mL = 209.04 \, g\)

02

Convert mass into moles by using molar mass

The molar mass of HSiCl3 is the sum of the atomic masses of its constituent atoms, which equals 1 (H) + 28.08 (Si) + 35.45 (Cl) x 3 = 170.88 g/mol. Thus, 209.04 g equals \( \frac{209.04 \, g}{170.88 \, g/mol} = 1.223 \, moles \) of HSiCl3.

03

Determine the mole ratio from the balanced reaction equation

Looking at the first reaction, the stoichiometric ratio of Si to HSiCl3 is 1:1. Hence, one should theoretically be able to obtain 1.223 moles of HSiCl3 per mole of Si.

04

Estimate the number of moles of HCl used

Using the Ideal Gas Law (PV=nRT), one can calculate the number of moles of HCl present, knowing that Pressure (P) = 10 atm, Volume (V) = 15 L, and Temperature (T) = 35℃ = 308.15 K. As for R = 0.08206 L·atm/K·mol, we get n= \( \frac{PV}{RT} = \frac{10 \, atm \times 15 \, L}{0.08206 \, L·atm/K·mol \times 308.15 \, K} = 59.01 \, mol \) of HCl.

05

Calculate the Theoretical Yield of HSiCl3

Looking at the first reaction, the stoichiometric ratio of HCl to HSiCl3 is 3:1. Hence, one should theoretically be able to obtain 59.01/3 = 19.67 moles of HSiCl3 per mole of HCl.

06

Determine the Percent yield

The percent yield of HSiCl3 is then given by \(\frac{actual \, yield}{theoretical \, yield} \times 100\% = \frac{1.223 \, mol}{19.67 \, mol} \times 100\% ≈ 6.22\%\) Phase 2: Determining the Volume of SiH4

07

Consider the stoichiometric ratio

In the second reaction, the stoichiometric ratio of HSiCl3 to SiH4 is 4:1. Therefore, it is expected that 0.25 moles of SiH4 would be obtained for each mole of HSiCl3.

08

Calculate the theoretical yield

Given that we have started with 1.223 moles of HSiCl3, we would expect \(1.223 \, mol \times 0.25 = 0.306 \, mol\) of SiH4.

09

Calculate the actual yield

The percent yield of SiH4 is 93.1%. Consequently, the actual yield is \(0.93 \times 0.306 = 0.285 \, mol\).

10

Use Ideal Gas Law to find the volume

Using the Ideal Gas Law (PV=nRT), one can calculate the volume of SiH4. The pressure (P) = 10 atm, the number of moles equals 0.285 mol, Temperature (T) = 35℃ = 308.15 K and R = 0.08206 L·atm/K·mol. We get \(Volume = \frac{nRT}{P} = \frac{0.285 \, mol \times 0.08206 \, L·atm/K·mol \times 308.15 \, K}{10 \, atm} = 7.11 \, L\). So, the volume of SiH4 obtained will be approximately 7.11 liters.

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