Question

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 \(93.1 \% ?\)

Short answer

To summarize, for part (a) the percent yield of HSiCl3 is approximately 6.22%. For part (b), the volume of SiH4 obtained will be approximately 7.11 liters under the given conditions, with a 93.1% yield of the reaction.

Step-by-step solution

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 \(156 \, mL \times 1.34 \, g/mL = 209.04 \, g\)

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.

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.

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.

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.

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

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.

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.

Calculate the actual yield

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

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.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us predict the quantities of reactants and products in a given chemical reaction. It relies heavily on the balanced chemical equation, which shows the proportions of reactants and products, known as stoichiometric coefficients. These coefficients tell us the mole ratio between different substances involved in the reaction.

In the provided exercise, stoichiometry plays a key role in determining how much HSiCl3 can be produced from a known amount of Si and HCl. Using the balanced equation, we see that for every mole of Si used, one mole of HSiCl3 is produced, making the mole ratio 1:1. Similarly, in the second reaction, the 4:1 ratio between HSiCl3 and SiH4 helps us understand how much SiH4 can be synthesized from a given amount of HSiCl3.

By understanding these ratios, we can accurately predict the theoretical yield of a reaction, which is essential for calculating the efficiency of the production process, also known as percent yield.

Ideal Gas Law

The Ideal Gas Law is an essential tool in chemistry that helps us calculate the properties of gases under various conditions. The law is summarized by the equation \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for the number of moles, \( R \) for the gas constant, and \( T \) for temperature in Kelvin. This equation allows us to solve for any of these variables when the others are known, making it incredibly useful for predicting the behavior of gases.

In our problem, the Ideal Gas Law is used to determine the number of moles of HCl gas available for the reaction and later to calculate the volume of produced SiH4 gas. Knowing the pressure, volume, and temperature of the gas, we can easily find its mole quantity, which is necessary for the stoichiometric calculations.

It is important to have all units matched appropriately: pressure should be in atm, volume in liters, and temperature in Kelvin, as the gas constant \( R \) is typically given as \( 0.08206 \) L·atm·K^-1·mol^-1. This consistency ensures accurate calculations, making the Ideal Gas Law a reliable method for dealing with gaseous reactants and products.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, calculated by comparing the actual yield to the theoretical yield. The formula to calculate percent yield is:

\[ \text{Percent Yield} = \left ( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right ) \times 100\% \]

In simpler terms, it tells us how much of the intended product we actually obtained from the reaction compared to the maximum possible amount we could have obtained. This efficiency gauge is vital, especially in industrial processes where the goal is to produce the most product from given reactants.

In the exercise, after calculating the theoretical yield of HSiCl3 based on stoichiometry, the actual yield is obtained from the experimental data. With these two values, we compute the percent yield, which in this scenario turned out to be 6.22%.

Knowing the percent yield helps chemists improve reaction conditions to increase the efficiency and optimize resource usage. It can also offer insights into any potential side reactions or losses that might be occurring.