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What is the percent yield of a reaction that produces 12.5 g CF2Cl2 from 32.9 g CCl4 and excess HF? CCl4+2HF⟶CF2Cl2+2HCl

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To calculate the percent yield of a reaction that produces CF2Cl2 from CCl4 and HF, you need to compare the actual yield to the theoretical yield. Convert the masses of CCl4 and CF2Cl2 to moles, use the balanced equation to determine the theoretical yield, and divide the actual yield by the theoretical yield to calculate the percent yield.

To calculate the percent yield, you need to compare the actual yield (the amount of CF2Cl2 obtained) to the theoretical yield (the amount of CF2Cl2 that would be obtained if the reaction went to completion).

First, convert the masses of CCl4 and CF2Cl2 to moles using their molar masses.

Next, use the stoichiometry of the balanced equation to determine the theoretical yield of CF2Cl2 from CCl4. Finally, divide the actual yield by the theoretical yield and multiply by 100 to get the percent yield.

The molar masses of CCl4 and CF2Cl2 are:

CCl4: 12.01 + 4(35.45) = 153.82 g/mol

CF2Cl2: 12.01 + 2(19.00) + 2(35.45) = 121.00 g/mol

Converting the masses to moles:

moles of CCl4 = 32.9 g / 153.82 g/mol = 0.214 mol

moles of CF2Cl2 = 12.5 g / 121.00 g/mol = 0.103 mol

Using stoichiometry, we can see that 1 mol of CCl4 produces 1 mol of CF2Cl2. Therefore, the theoretical yield of CF2Cl2 from CCl4 is also 0.214 mol.

Finally, calculating the percent yield:

Percent yield = (actual yield / theoretical yield) * 100 = (0.103 mol / 0.214 mol) * 100 = 48.13%

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