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

Given equation is
2NaCl + `H_(2)SO_(4)to Na_(2)SO_(4)+2HCl`
1 Mole `H_(2)SO_(4)to 2` Moles HCl
98 gms of `H_(2)SO_(4)to 2 xx 36.5 =73` gms of HCl ?
`to 420 xx 10^(3)` gms of HCl
`(420x10^(3)xx98)/(73)=563.84xx10^(3)` gms
For 100 % `H_(2)SO_(4)to 563.84 x x10^(3)` gms
For 90% `H_(2)SO_(4)to 563.84xx10^(3)xx(100)/(90)`
= 626.5 xx `10^(3)"gms of" H_(2)SO_(4)`
`:. 626.5 xx 10^(3)` gms of 90% `H_(2)SO_(4)` is required to produce 420 x x `10^(3)` gms of HCl