Answer: The percentage yield of iron sulfide for the reaction is 69.04 %.
Explanation:
For the given chemical equation:
\(2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr\)
To calculate the percentage yield of a compound, we use the following equation:
\(\%\text{ yield}=\frac{\text{Measured yield}}{\text{Calculated yield}}\times 100\)
We are given:
Measured yield of iron sulfide = 11.6 g
Calculated yield of iron sulfide = 16.8 g
Putting values in above equation, we get:
\(\%\text{ yield of iron sulfide}=\frac{11.6g}{16.8g}\times 100=69.04\%\)
Hence, the percentage yield of iron sulfide for the reaction is 69.04 %.
