Calculate the amount of 90% `H_(2)SO_(4)` required for the preparation of 420 kg HCl.
`2NaCl+H_(2)SO_(4)toNa_(2)SO_(4)+2HCl`

Explanation

No. of moles of HCl to be prepared
`=(420xx10^(3))/36.5=11.5xx10^(3)`
According to the reaction for every two moles of HCl one mole of `H_(2)SO_(4)` required is
`11.5/2xx10^(3)=5.75xx10^(3)`
Wt. of `H_(2)SO_(4)=5.75xx10^(3)xx98=563.5kg`
Since the given `H_(2)SO_(4)` contain only 90%.
The weight to be taken is `(563.5xx100)/90=627kg`.