Questions: What is the percent yield of iron chloride if we start with 34 g iron bromide and produce 4 g iron chloride? FeBr2 + 2 KCl → FeCl2 + 2 KBr Select the correct answer below: 11.39 % 18.37 % 20.02 % 37.48 %

What is the percent yield of iron chloride if we start with 34 g iron bromide and produce 4 g iron chloride?
FeBr2 + 2 KCl → FeCl2 + 2 KBr

Select the correct answer below:
11.39 %
18.37 %
20.02 %
37.48 %

Transcript text: What is the percent yield of iron chloride if we start with 34 g iron bromide and produce 4 g iron chloride? $\mathrm{FeBr}_{2}+2 \mathrm{KCl} \rightarrow \mathrm{FeCl}_{2}+2 \mathrm{KBr}$ Select the correct answer below: $11.39 \%$ $18.37 \%$ $20.02 \%$ $37.48 \%$

Solution

Solution Steps

Step 1: Calculate the Molar Mass of Iron Bromide and Iron Chloride

First, we need to calculate the molar masses of iron bromide ($\mathrm{FeBr}_2$) and iron chloride ($\mathrm{FeCl}_2$).

Molar mass of $\mathrm{FeBr}_2$:
- Iron (Fe): $55.845 \, \text{g/mol}$
- Bromine (Br): $79.904 \, \text{g/mol}$
- Molar mass of $\mathrm{FeBr}_2 = 55.845 + 2 \times 79.904 = 215.653 \, \text{g/mol}$
Molar mass of $\mathrm{FeCl}_2$:
- Iron (Fe): $55.845 \, \text{g/mol}$
- Chlorine (Cl): $35.453 \, \text{g/mol}$
- Molar mass of $\mathrm{FeCl}_2 = 55.845 + 2 \times 35.453 = 126.751 \, \text{g/mol}$

Step 2: Calculate the Theoretical Yield of Iron Chloride

Using stoichiometry, we calculate the theoretical yield of $\mathrm{FeCl}_2$ from 34 g of $\mathrm{FeBr}_2$.

Convert 34 g of $\mathrm{FeBr}_2$ to moles: \[ \text{Moles of } \mathrm{FeBr}_2 = \frac{34 \, \text{g}}{215.653 \, \text{g/mol}} = 0.1576 \, \text{mol} \]
According to the balanced equation, 1 mole of $\mathrm{FeBr}_2$ produces 1 mole of $\mathrm{FeCl}_2$. Therefore, the moles of $\mathrm{FeCl}_2$ produced is also 0.1576 mol.
Convert moles of $\mathrm{FeCl}_2$ to grams: \[ \text{Theoretical yield of } \mathrm{FeCl}_2 = 0.1576 \, \text{mol} \times 126.751 \, \text{g/mol} = 19.975 \, \text{g} \]

Step 3: Calculate the Percent Yield

The percent yield is calculated using the actual yield and the theoretical yield:

\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% \]

Given that the actual yield is 4 g:

\[ \text{Percent yield} = \left( \frac{4 \, \text{g}}{19.975 \, \text{g}} \right) \times 100\% = 20.02\% \]