Calculate the volume of hydrogen librated at N.T.P. when `500 cm^(3)` of `0.5 N` sulphuric acid react with excess of zinc.

Answer

`2800 cm^(3)`

Explanation

Step I. Mass of sulphuric acid in solution
`"Normality" (N) = (("Mass of" H_(2)SO_(4))/("Equivalent mass"))/(("Volume of solution in" cm^(3))/(100))` or `(0.5 "equiv" dm^(-3)) = (W)/(((49 "equiv"^(-1)))/((0.5 dm^(3))))`
`W = (0.5 "equiv" dm^(-3)) xx (49 g "equiv"^(-1)) xx (0.5 dm^(3)) = 12.25 g`.
Step II. Volume of hydrogen liberarted at N.T.P.
The chemical equation for the reaction is :
`{:(Zn+H_(2)SO_(4),rarrZnSO_(4)+,H_(2)),(2+32+64,,22400cm^(3)),(=98g,,):}`
`98g` of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P. `22400 cm^(3)`
`12.25 g` of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P. `= 22400 (cm^(3)) xx ((12.25 g))/((98g)) = 2800 cm^(3)`