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

`HNO_(3)` oxidies `NH_(4)^(+)` ions to nitrogen and itself gets reduce

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

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

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