Question

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\(\% ?\)

Short answer

= 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.

Step-by-step solution

Calculate the number of moles of HCl used

We are given 15.0 L of HCl gas at 10.0 atm and 35 degrees Celsius. We can use the ideal gas law to find the number of moles of HCl: PV = nRT where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature (in K). First, we need to convert the temperature to Kelvin: T(K) = 35 + 273.15 = 308.15 K Now, we can solve for n: n = PV/(RT) = (10.0 atm)(15.0 L) / (0.0821 L atm/mol K)(308.15 K)

Key concepts

Ideal Gas Law

The ideal gas law, one of the cornerstone equations in chemistry, helps us understand the behavior of gases under different conditions. It's written as \( PV = nRT \), where:

• \( P \) is the pressure of the gas (measured in atmospheres or atm).
• \( V \) is the volume of the gas (measured in liters or L).
• \( n \) represents the number of moles of the gas.
• \( R \) is the ideal gas constant, valued at \( 0.0821 \) L atm/mol K.
• \( T \) is the temperature of the gas (measured in Kelvin or K).

To solve problems using the ideal gas law, you first need to ensure all units are correct. For example, temperatures should be converted to Kelvin by adding 273.15 to the Celsius value.

In the exercise, this law was used to determine how many moles of \( \text{HCl} \) were involved. By rearranging the formula to solve for \( n \), the equation becomes \( n = \frac{PV}{RT} \). Substituting the given values allows you to calculate the moles of the gas involved in the reaction.

Percent Yield

Percent yield is a concept that measures the efficiency of a chemical reaction by comparing the actual yield to the theoretical yield. It is particularly useful when assessing the productivity of industrial chemical processes. The formula for calculating the percent yield is:

• Percent Yield = \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \)

The actual yield is the quantity of product actually obtained from an experiment or industrial process, while the theoretical yield is the amount expected based on stoichiometry, the balanced chemical equations.

In our exercise problem, this concept was used to gauge the synthesis of \( \text{HSiCl}_3 \). After calculating the actual moles obtained and comparing it with the predicted moles (from the ideal gas calculations), you apply these values to the formula. A percent yield above 100% is not possible, indicating errors in measurement or process.

Chemical Reactions

Understanding the way in which reactants transform into products during a chemical reaction is a pivotal part of chemistry. In our exercise, we dealt with two reactions:

• \( \text{Si}(s) + 3 \text{HCl}(g) \rightarrow \text{HSiCl}_3(l) + \text{H}_2(g) \)
• \( 4 \text{HSiCl}_3(l) \rightarrow \text{SiH}_4(g) + 3 \text{SiCl}_4(l) \)

Chemical equations like these provide basic information about the reactants and products involved, as well as the stoichiometric ratios necessary for calculations.

Stoichiometry, the study of these mass and molar relationships, allows for predictions about quantities of substances consumed or produced. For example, for every mole of \( \text{Si} \), three moles of \( \text{HCl} \) are required. In turn, these stoichiometric ratios help us calculate actual yields and assess the feasibility and efficiency of reactions in both laboratory and industrial settings.