A colourless salt X has 50% `Na_(2)SO_(3)` and 50% `H_(2)O` . How much of `SO_(2)` at NTP is obtained when 2.52 g of X reacts with excess of dil . `H_(2)SO_(4)` ?

A. 22.4 L

B. 0.448 L

C. 44.8 L

D. 0.224 L

A. 22.4 L

B. 0.448 L

C. 44.8 L

D. 0.224 L

Correct Answer – D

In salt (X)=`underset(50%)(Na_(2)SO_(3)):underset(50%)(H_(2)O)`

Molar mass of the `Na_(2)SO_(3)=126`

Molar mass of the `H_(2)O=18`

Hence , 7 molecules of water are attached with `Na_(2)SO_(3)`

`therefore ` Formula =`Na_(2)SO_(3)*7H_(2)O`

`126:18xx 7 implies 126:126implies 1:1`

`Na_(2)SO_(3)*7H_(2)O + H_(2)SO_(4)(aq) to SO_(2)(aq) + Na_(2)SO_(3)(aq) + H_(2)O(l)`

`because 252 g Na_(2)SO_(3)* 7H_(2)O ` gives `=22.4L SO_(2)`

`therefore 2.52 g Na_(2)SO_(3) *7H_(2)O` will give `=(22.4)/(252)xx252=0.224 L`