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

Step-by-step 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 =
`(64)/(68.2)* 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 =
`(96)/(68.2)* 7.5=10.5g` of
`S`

% yield=
`\frac{\text{Actual yield}}{\text {Theoretical yield}}* 100=(7.95g)/(10.5g)* 100=75.7\%`

Thus 75.7 % yield is there.