Correct Answer - C
`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.