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? $$\mathrm{Si}(\mathrm{s})+2 \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g}) \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Si} \mathrm{Cl}_{2}(\mathrm{g})$$ What pressure of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})\) would you expect in this same flask at \(95^{\circ}\) 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.)

Short answer

13.35 g of dimethyldichlorosilane can be formed, and the pressure would be about 360.24 mm Hg.

Step-by-step solution

Convert Pressure to Moles of CH3Cl

First, convert the given pressure of CH₃Cl to moles using the ideal gas law: \[ PV = nRT \]where:- \( P = 585 \) mm Hg, which is equivalent to \( 585 \times \frac{1}{760} \) atm,- \( V = 6.56 \) L,- \( R = 0.0821 \) L atm / (K mol),- \( T = 25 + 273.15 = 298.15 \) K.Plug in the values to find \( n \):\[ n = \frac{PV}{RT} = \frac{(585/760) \times 6.56}{0.0821 \times 298.15} \approx 0.206 \text{ moles CH}_3\text{Cl} \]

Identify the Limiting Reactant

The stoichiometry of the reaction \( \text{Si}( ext{s}) + 2 \text{CH}_3 \text{Cl} \rightarrow (\text{CH}_3)_2\text{SiCl}_2(\text{g}) \) shows that 2 moles of CH₃Cl react with 1 mole of Si to produce 1 mole of dimethyldichlorosilane.Calculate moles of Si:\[ \text{Molar mass of Si} = 28.09 \text{ g/mol}, \text{Moles of Si} = \frac{2.25}{28.09} \approx 0.0801 \text{ moles Si} \]Check limiting reactant:- For 1 mole of Si, 2 moles of CH₃Cl are required, but only have 0.206 moles CH₃Cl.- Moles of CH₃Cl needed for 0.0801 mol Si: \( 0.0801 \times 2 = 0.1602 \text{ moles CH}_3\text{Cl} \)Si is not the limiting reactant; CH₃Cl is, as all 0.206 moles of it will be consumed.

Calculate Mass of Dimethyldichlorosilane Formed

Since CH₃Cl is the limiting reactant:- 0.206 moles CH₃Cl produces \( \frac{0.206}{2} = 0.103 \) moles of \((\text{CH}_3)_2\text{SiCl}_2\).Find the mass of \((\text{CH}_3)_2\text{SiCl}_2\):- Molar mass of \((\text{CH}_3)_2\text{SiCl}_2\) = \((12.01 \times 2 + 1.01 \times 6 + 28.09 + 35.45 \times 2) \approx 129.62\) g/mol.- Mass produced: \[ 0.103 \text{ moles} \times 129.62 \text{ g/mol} = 13.35 \text{ g} \]

Calculate Pressure of Dimethyldichlorosilane at 95°C

Now calculate the pressure of \((\text{CH}_3)_2\text{SiCl}_2\) at 95°C (368.15 K) using the ideal gas law and the volume remains constant:Number of moles of \((\text{CH}_3)_2\text{SiCl}_2\) = 0.103\[ P = \frac{nRT}{V} = \frac{0.103 \times 0.0821 \times 368.15}{6.56} \approx 0.474 \text{ atm} \]Convert pressure back to mm Hg:\[ 0.474 \times 760 = 360.24 \text{ mm Hg} \]

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely used up first, thereby determining the maximum amount of product that can be produced. When one reactant is entirely consumed, the reaction comes to a halt even if other reactants are left behind. This concept is vital because it helps predict the amounts of products formed, preventing waste and optimizing efficiency.

To identify the limiting reactant, compare the mole ratios of the reactants involved in the chemical equation. In our example, according to the balanced equation:

• 0.0801 moles of Si is available, requiring 0.1602 moles of CH₃Cl.
• However, we only have 0.206 moles of CH₃Cl.

Since 0.206 moles is greater than 0.1602 moles, CH₃Cl becomes the limiting reactant. Recognizing this helps us determine that the reaction stops once all of CH₃Cl is consumed, hence determining the amount of dimethyldichlorosilane produced.

Chemical Stoichiometry

Chemical stoichiometry provides a quantitative relationship between reactants and products in a chemical reaction, based on the balanced equation. It allows us to predict how much of each substance is involved in a reaction.

In the equation \[ \text{Si}(\text{s}) + 2 \text{CH}_3 \text{Cl} \rightarrow (\text{CH}_3)_2 \text{SiCl}_2(\text{g}) \] The coefficients denote that one mole of Si reacts with two moles of CH₃Cl to produce one mole of dimethyldichlorosilane. This molar relationship guides the stoichiometric calculations:

• If CH₃Cl is the limiting reactant, its amount dictates the amount of product formed.
• Here, 0.206 moles of CH₃Cl yields half as many moles of dimethyldichlorosilane, aligning with its stoichiometric coefficient.

Chemical stoichiometry is instrumental in determining these precise amounts, ensuring reactions are carried out efficiently without excess leftover reactants.

Molar Mass Calculation

Calculating molar mass plays a pivotal role in converting between grams and moles, crucial for stoichiometry in chemical reactions. It involves summing the atomic masses of all atoms in a molecule.

For example, the molar mass of Si is: \[ 28.09 \text{ g/mol} \] For dimethyldichlorosilane, \((\text{CH}_3)_2\text{SiCl}_2\), we calculate its molar mass as follows:

• Each carbon (C) atom contributes 12.01 g/mol: \(2 \times 12.01 = 24.02\text{ g/mol}\).
• Hydrogen (H) atoms contribute 1.01 g/mol: \(6 \times 1.01 = 6.06\text{ g/mol}\).
• Silicon (Si) contributes 28.09 g/mol.
• Each chlorine (Cl) atom contributes 35.45 g/mol: \(2 \times 35.45 = 70.90\text{ g/mol}\).

Adding these gives the total molar mass: \[ 129.62 \text{ g/mol} \] This is then used to convert moles of product to grams, aiding in practical measures of substance production, such as finding the 13.35 g of dimethyldichlorosilane produced in the reaction.