Question

The percent yield for the following reaction carried out in carbon tetrachloride (CCI4) solution is 80%

$Br_2 + Cl_2 \longrightarrow 2BrCl$

(a) How many moles of BrCl is formed from the reaction of 0.025 mol $Br_2$ and 0.025 mol $Cl_2$?

(b) How many moles of $Br_2$ is left unreacted?

Answer 1

(a) Moles of BrCl formed: 0.040 mol

(b) Moles of $Br_2$ left unreacted: 0.005 mol

Answer 2

1. Write the balanced equation: $Br_2 + Cl_2 \longrightarrow 2BrCl$.

2. Identify initial moles: $Br_2 = 0.025$ mol, $Cl_2 = 0.025$ mol.

3. Since the initial moles are in the stoichiometric ratio (1:1), both reactants are in stoichiometric proportion.

4. Calculate the theoretical yield of BrCl: Based on 0.025 mol of $Br_2$ (or $Cl_2$), the theoretical yield is $0.025 \times 2 = 0.050$ mol BrCl.

5. Calculate the actual yield using the percent yield: Actual yield = $0.050 \text{ mol} \times 0.80 = 0.040$ mol BrCl. (Answer to part a)

6. Calculate the moles of $Br_2$ consumed based on the actual yield: Moles consumed = $0.040 \text{ mol BrCl} \times \frac{1 \text{ mol } Br_2}{2 \text{ mol BrCl}} = 0.020$ mol $Br_2$.

7. Calculate the moles of $Br_2$ left unreacted: Moles left = Initial moles - Moles consumed = $0.025 \text{ mol} - 0.020 \text{ mol} = 0.005$ mol. (Answer to part b)