Question

For the following reaction, 3.59 grams of carbon tetrachloride are mixed with excess methane (CH4) . Assume that the percent yield of dichloromethane (CH2Cl2) is 65.5 %.What is the ideal yield of dichloromethane (CH2Cl2) ? What is the actual yield of dichloromethane (CH2Cl2)

Answer 1

The balanced chemical equation for the reaction is:

CCl4 + CH4 → 2CH2Cl2

From the balanced equation, we see that 1 mole of CCl4 reacts with 1 mole of CH4 to produce 2 moles of CH2Cl2.

To determine the theoretical yield of CH2Cl2, we need to convert the given mass of CCl4 to moles, using its molar mass:

Molar mass of CCl4 = 12.01 + 4(35.45) = 154.02 g/mol

Moles of CCl4 = 3.59 g / 154.02 g/mol = 0.0233 mol

Since 1 mole of CCl4 produces 2 moles of CH2Cl2, the theoretical yield of CH2Cl2 is:

Theoretical yield of CH2Cl2 = 2 × 0.0233 mol = 0.0466 mol

To determine the actual yield of CH2Cl2, we need to use the percent yield:

Percent yield = (actual yield / theoretical yield) × 100%

Rearranging this formula gives:

Actual yield = (percent yield / 100%) × theoretical yield

Substituting the given values, we get:

Actual yield = (65.5% / 100%) × 0.0466 mol = 0.0305 mol

Finally, we can convert the actual yield of CH2Cl2 to grams, using its molar mass:

Molar mass of CH2Cl2 = 12.01 + 2(35.45) = 84.93 g/mol

Mass of CH2Cl2 = 0.0305 mol × 84.93 g/mol = 2.59 g

Therefore, the ideal yield of CH2Cl2 is 0.0466 mol or 3.97 g, and the actual yield is 0.0305 mol or 2.59 g.