Answer: The percentage yield of iron sulfide for the reaction is 69.04 %.
Explanation:
For the given chemical equation:
[tex]2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr[/tex]
To calculate the percentage yield of a compound, we use the following equation:
[tex]\%\text{ yield}=\frac{\text{Measured yield}}{\text{Calculated yield}}\times 100[/tex]
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:
[tex]\%\text{ yield of iron sulfide}=\frac{11.6g}{16.8g}\times 100=69.04\%[/tex]
Hence, the percentage yield of iron sulfide for the reaction is 69.04 %.
