Question

How many grams of silver nitrate are required to produce 0.25 mole of silver sulphide

Answer 1

Given: 0.25 mole of silver sulphide.

To find: We have to find the amount of silver nitrate is required to produce 0.25 mole of silver sulphide.

Solution:

The chemical formula of silver sulfide is $Ag_2S$.

Silver nitride reacts with hydrogen sulfide to produce silver sulphide along with nitric acid.

The reaction is-

$2AgNO_3+H_2S\rightarrow Ag_2S+2HNO_3$.

So, 1 mole of silver sulphide is produced from 2 moles of silver nitrate.

0.25 mole of silver sulphide is produced from 2×0.25=0.5 moles of silver nitrate.

The molar mass of silver nitrate is 170.

Thus the amount of silver nitrate required is 170×0.5=85 gram.

85 grams of silver nitrate is required to produce 0.25 mole of silver sulphide.