Question

The reaction of 11.9 g of chcl3 with excess chlorine produced 11.2 g of ccl4, carbon tetrachloride: what is the percent yield?

Answer 1

73.0% First, lookup the atomic weights of the elements involved Atomic weight carbon = 12.0107 Atomic weight chlorine = 35.453 Atomic weight hydrogen = 1.00794 Calculate the molar masses. Molar mass CHCl3 = 12.0107 + 1.00794 + 3 * 35.453 = 119.37764 Molar mass CCl4 = 12.0107 + 4 * 35.453 = 153.8227 Calculate the number of moles we have Moles CHCl3 = 11.9 g / 119.37764 g/mol = 0.099683659 mol Moles CCl4 = 11.2 g / 153.8227 g/mol = 0.0728111 mol Since with 100% yield, each molecule of CHCl3 should produce one molecule of CCl4, divide the actual moles produced by the number of moles expected for 100% yield. So 0.0728111 / 0.099683659 = 0.730421623 = 73.0421623 % Rounding to 3 significant figures gives 73.0%

Answer 2

First, we need to calculate the theoretical yield as follows:
The balanced chemical reaction is:
2CHCl3 +2Cl2 ----> 2CCl4 +2HCl
molar mass of CHCl3 = 119.378 grams
molar mass of CCl4 = 153.823 grams
number of moles = mass / molar mass = 11.9 / 119.378 = 0.0997 moles
From the balanced equation, 2 moles of CHCl3 produce 2 moles of CCl4.
Therefore, 0.0997 moles of CHCl4 will produce 0.0997 moles of CCl4.
number of moles = mass / molar mass
mass (theoretical) = molar mass * number of moles
mass (theoretical) = 153.823 * 0.0997 = 15.336 grams

Now, we will calculate the percent yield as follows:
% yield = (actual mass / theoretical mass) * 100
% yield = (11.2 / 15.336) * 100 = 73.03%