Problem 60 - Chemistry and Chemical Reactivity

Question

If you place \(2.25 \mathrm{g}\) of solid silicon in a \(6.56-\mathrm{L}\). flask that contains \(\mathrm{CH}_{3} \mathrm{Cl}\) with a pressure of \(585 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\) what mass of dimethyldichlorosilane, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})\) can be formed? $$\mathrm{Si}(\mathrm{s})+2 \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g}) \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})$$ What pressure of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})\) would you expect in this same flask at \(95^{\circ} \mathrm{C}\) on completion of the reaction? (Dimethyldichlorosilane is one starting material used to make silicones, polymeric substances used as lubricants, antistick agents, and water-proofing caulk.)

Step-by-Step Solution

Answer

10.35 g of product formed; the pressure is 282 mmHg at 95°C.

Step 1: Identify Given Information

We have the initial conditions of 2.25 g of solid Si and \(\mathrm{CH}_{3} \mathrm{Cl}\) gas with a pressure of 585 mm Hg at 25°C. The reaction is given by \[ \mathrm{Si(s)} + 2 \mathrm{CH}_{3} \mathrm{Cl(g)} \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(g) \]. We need to calculate the mass of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(g) \) formed and the pressure at 95°C.

Step 2: Calculate Moles of Si

The molar mass of Si is approximately 28.09 g/mol. Calculate moles of Si from its mass: \[ \text{Moles of } \mathrm{Si} = \frac{2.25\, \mathrm{g}}{28.09\, \mathrm{g/mol}} \approx 0.0802\, \mathrm{mol} \]

Step 3: Convert Pressure to Atm

To use the ideal gas law, convert the pressure of \(\mathrm{CH}_{3} \mathrm{Cl}\) from mm Hg to atm: \[ 585\, \mathrm{mmHg} \times \frac{1 \mathrm{atm}}{760 \mathrm{mmHg}} \approx 0.770\, \mathrm{atm} \]

Step 4: Calculate Moles of CH3Cl

Use the ideal gas law to find moles of \(\mathrm{CH}_{3} \mathrm{Cl}\): \[ PV = nRT \Rightarrow n = \frac{PV}{RT} \] Given \(P = 0.770\, \mathrm{atm}\), \(V = 6.56\, \mathrm{L}\), \(R = 0.0821\, \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}\), \(T = 298\, \mathrm{K}\): \[ n = \frac{(0.770)(6.56)}{(0.0821)(298)} \approx 0.207\, \mathrm{mol} \]

Step 5: Determine Limiting Reactant

For the reaction \( \mathrm{Si(s)} + 2 \mathrm{CH}_{3} \mathrm{Cl(g)} \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(g) \), \(0.207\) mol \(\mathrm{CH}_{3} \mathrm{Cl}\) requires \(0.1035\, \text{mol Si}\), hence \(\mathrm{Si}\) is the limiting reactant.

Step 6: Calculate Mass of Product Formed

The molar mass of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}\) is approximately 129.08 g/mol. The reaction of 0.0802 mol of \(\mathrm{Si}\) produces equal moles of product: \[ \mathrm{Mass} = 0.0802\, \mathrm{mol} \times 129.08\, \mathrm{g/mol} \approx 10.35\, \mathrm{g} \]

Step 7: Calculate Pressure at New Temperature

Using the ideal gas law again to find the pressure at 95°C (368 K): \(n = 0.0802 \) mol, \(V = 6.56 \mathrm{L}\), \(R = 0.0821 \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}\), \(T = 368 \mathrm{K}\): \[ P = \frac{nRT}{V} = \frac{(0.0802)(0.0821)(368)}{6.56} \approx 0.371\, \mathrm{atm} \]

Step 8: Final Step: Convert Pressure to mmHg

Convert the pressure of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(g)\) from atm to mmHg: \[ 0.371\, \mathrm{atm} \times 760\, \mathrm{mmHg/atm} \approx 282\, \mathrm{mmHg} \]

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of an ideal gas in terms of pressure, volume, temperature, and moles of gas present. This law is articulated as: \[PV = nRT\] Where:

• \(P\) is the pressure of the gas in atmospheres (atm).
• \(V\) is the volume of the gas in liters (L).
• \(n\) is the number of moles of gas.
• \(R\) is the ideal gas constant, approximately \(0.0821 \, \text{L} \cdot \text{atm}/\text{mol} \cdot \text{K}\).
• \(T\) is the temperature in Kelvin (K), calculated as the Celsius temperature plus 273.15.

This relationship is crucial for determining unknown quantities, such as the moles of a gas present in a given volume or the pressure a gas would exert at different conditions. For example, in our exercise, the Ideal Gas Law is used to find the moles of \(\mathrm{CH}_3 \mathrm{Cl}\) in the flask based on its pressure and volume. Then it helps to predict the pressure of the product gas, dimethyldichlorosilane, at a new temperature after the reaction occurs. To apply the Ideal Gas Law effectively:

• Ensure units match (convert temperatures to Kelvin and pressures to atm when necessary).
• Remember that \(R\) remains constant for all ideal gases.
• Make use of the equation to solve for the unknown variable when other values are provided.

By understanding this law, students can predict how changing one property of a gas affects the others, showcasing its power in chemical calculations.

Stoichiometry

Stoichiometry is the calculation of the relative quantities of reactants and products in chemical reactions. It ensures that the Law of Conservation of Mass is satisfied and is key in making precise predictions about how much product can be formed from given reactants. In a balanced chemical equation, stoichiometric coefficients indicate the proportion of substances that react or form. For the reaction given:\[ \mathrm{Si(s)} + 2 \mathrm{CH}_3 \mathrm{Cl(g)} \rightarrow \left( \mathrm{CH}_3 \right)_2 \mathrm{SiCl}_2(g) \] The coefficients imply that 1 mole of \(\mathrm{Si}\) reacts with 2 moles of \(\mathrm{CH}_3 \mathrm{Cl}\) to form 1 mole of \(\left( \mathrm{CH}_3 \right)_2 \mathrm{SiCl}_2\). This proportionality is critical when determining how much of each substance is needed or produced. In our exercise, stoichiometry is used to relate 0.0802 moles of \(\mathrm{Si}\) to the quantities of other substances. This is crucial for calculating the eventual mass of the product formed. Important tips in stoichiometry:

• Always ensure the chemical equation is balanced.
• Use the mole ratio (from coefficients) to convert between moles of different substances.
• Remember that mass and moles are related via molar mass, allowing conversion between grams and moles.

Mastery of stoichiometry enables accurate predictions about chemical processes and is a cornerstone of chemistry.

Limiting Reactant

The concept of the limiting reactant is essential in predicting the amount of product formed in a chemical reaction. The limiting reactant is the substance that is completely consumed in a reaction and thus determines the maximum amount of product that can be formed. To identify the limiting reactant, compare the mole ratio of the reactants provided to the balanced equation ratio. For the reaction:\[ \mathrm{Si(s)} + 2 \mathrm{CH}_3 \mathrm{Cl(g)} \rightarrow \left( \mathrm{CH}_3 \right)_2 \mathrm{SiCl}_2(g) \] You can calculate the moles of each reactant. In our exercise, we calculated:

• Moles of \(\mathrm{Si}\): 0.0802 mol
• Moles of \(\mathrm{CH}_3 \mathrm{Cl}\): 0.207 mol

According to the stoichiometry, \(1\) mole of \(\mathrm{Si}\) would require \(2\) moles of \(\mathrm{CH}_3 \mathrm{Cl}\), totaling \(0.0802 \times 2 = 0.1604\) moles of \(\mathrm{CH}_3 \mathrm{Cl}\). Since we have more than this required amount, \(\mathrm{Si}\) is the limiting reactant. Understanding the limiting reactant:

• Allows for the calculation of the theoretical yield - the maximum amount of product expected based on complete consumption of the limiting reactant.
• Is crucial for efficient utilization of resources in chemical manufacturing and lab experiments.

By grasping the concept of the limiting reactant, students can effectively analyze reaction completeness and efficiency, ensuring resource optimization.