Question

2). H2S and SO2 as follows,(8pt)

H2S + SO2 ------>S + H2O

In a particular experiment, 7.95 g of sulfur is produced by the reaction of 7.5 g of H2S with 12.75 g of SO2.


What is the % yield of sulfur?

Answer 1

Answer: 75.7 % yield

Explanation:

The balanced chemical equation is:

$2H_2S+SO_2\rightarrow 3S+2H_2O$  

According to stoichiometry :

68.2 g of $H_2S$ will require = 64 g of $SO_2$

Thus 7.5 g of $H_2S$ will require = $\frac{64}{68.2}\times 7.5=7.0g$ of $SO_2$

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 68.2 g of give = 96 g of

Thus 7.5 g of $H_2S$ give =$\frac{96}{68.2}\times 7.5=10.5g$  of $S$

% yield=$\frac{\text{Actual yield}}{\text {Theoretical yield}}\times 100=\frac{7.95g}{10.5g}\times 100=75.7\%$

Thus 75.7 % yield is there.