What weight of `HNO_(3)` is needed to convert 62gm of `P_(4)` in `H_(3) PO_(4) ` in the region ?
`P_(4) + HNO_(3) rarr H_(3) PO_(4) + NO_(2) + H_(2) O`

Options

A. 63 gm
B. 630gm
C. 315 gm
D. 126 gm

Answer

B

Explanation

The equiv. wt. of `P_(4) = ( 31 xx 4)/( 5 xx 4) = ( 31)/( 5)`
`:.` 62 gm `P_(4) = ( 62 xx 5)/( 31) ` equiv of `P_(4) = 10` equiv of `P_(4)`
The equiv. wt. of `HNO_(3)= ("Mol. wt.")/( 1) = ( 63)/( 1)`
`:.` the wt. of `HNO_(3)` required
`= 10 xx 163 = 630 gm`