`HNO_(3)` oxidies `NH_(4)^(+)` ions to nitrogen and itself gets reduced to `NO_(2)`. The moles of `HNO_(3)` required by 1 mole of `(NH_(4))_(2)SO_(4)` is:

Options

A. (a)`4`
B. (b)`5`
C. (c )`6`
D. (d)`2`

Answer

C

Explanation

`HNO_(3)+NH_(4)^(+) rarr N_(2)+NO_(2)`
V.F. of `HNO_(3)=(5-4)=1`
`V.F. of NH_(4)^(+)=[0-(-3)]=3`
so molar ratio of `HNO_(3)` and `NH_(4)^(+)` is `3:1`.
1 mole `(NH_(4))_(2)SO_(4)` is found to contain 2 mole of `NH_(4)^(+)`
So, required moles of `HNO_(3) is 3xx2=6` moles.