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For the following reaction, 3.69 grams of methane (\(CH_4\)) are mixed with excess carbon tetrachloride. The reaction yields 36.3 grams of dichloromethane (\(CH_2Cl_2\)).

\[ \text{methane (CH}_4\text{) (g) + carbon tetrachloride (CCl}_4\text{) (g) } \rightarrow \text{ dichloromethane (CH}_2Cl_2\text{) (g)} \]

1. What is the theoretical yield of dichloromethane (\(CH_2Cl_2\)) in grams?
2. What is the percent yield for this reaction?

Answer :

The theoretical yield of dichloromethane is 36.29 grams.

The percent yield is approximately 100.05%.

To calculate the theoretical yield and percent yield, follow these steps:

Calculate the Theoretical Yield:

Given that the balanced equation is:

CH₄(g) + CCl₄(g) → CH₂Cl₂(g)

Calculate the molar mass of CH₂Cl₂:

Molar mass of CH₂Cl₂ = 12.01 g/mol (C) + 2 * 1.01 g/mol (H) + 2 * 35.45 g/mol (Cl) = 84.93 g/mol

Calculate the theoretical moles of CH₂Cl₂ produced from 3.69 g of CH₄:

Moles of CH₂Cl₂ = Mass / Molar mass = 36.3 g / 84.93 g/mol ≈ 0.427 mol

Calculate the Percent Yield:

Theoretical yield = Moles of CH₂Cl₂ × Molar mass of CH₂Cl₂ = 0.427 mol × 84.93 g/mol ≈ 36.29 g

Percent Yield = (Actual yield / Theoretical yield) × 100

Percent Yield = (36.3 g / 36.29 g) × 100 ≈ 100.05%

Theoretical Yield of Dichloromethane (CH₂Cl₂): Approximately 36.29 grams

Percent Yield: Approximately 100.05%

The percent yield is slightly over 100%, which is not physically possible. This could be due to experimental errors, such as inaccuracies in measurements, losses during the reaction, or impurities. It's important to note that percent yield should ideally be less than 100%, indicating that the actual yield is lower than the theoretical yield.

To know more about dichloromethane refer here:

https://brainly.com/question/30715276#

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