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.

Explanation

The chemical equation representing the reaction is :
`Zn+underset(98.0g)(H_(2)SO_(4))rarrZnSO_(4)+underset(22400cm^(3)"at STP")(H_(2))`
Step 1. To calculate the amount of `H_(2)SO_(4)` in `500 cm^(3)` of 0.5 N `H_(2)SO_(4)` solution.
Strength/litre of `H_(2)SO_(4)" solution" = "Normality"xx"Eq. mass"=0.5xx49=24.5"g/litre"`
Now, `1000 cm^(3)` of the acid solution contain 24.5 g of pure `H_(2)SO_(4)`
`therefore 500cm^(3)" of the acid solution will contain pure "H_(2)SO_(4)=24.5xx(500)/(1000)=12.25g`
Step 2. to calculate the volume of `H_(2)` liberated at STP
`" 98 g of "H_(2)SO_(4)" react to liberate "22400cm^(3)" of "H_(2)" at STP"`
`therefore" 12.25 g of "H_(2)SO_(4)" would liberate H"_(2)" at STP"=(22400)/(98)xx12.25cm^(3)=2800cm^(3)`
Thus, the volume of hydrogen liberated at STP = `2800cm^(3)`