What quanitity of ammonium sulphate is necessary for the production of `NH_(3)` gas sufficient to neutralize a solution containing 292 g of HC1? `[HCl=36.5,(NH_(4))_(2) SO_(4)=132, NH_(3)=17]`

Options

A. 272 g
B. 403 g
C. 528 g
D. 1056 g

Answer

C

Explanation

1 moles of `NH_(3)` neutralizes one mole of HCl , Now, moles of `Hcl = (292)/(36.5) = 8`
`rArr` Moles of `NH_(3)` required `= 8 rArr` Moles of `(NH_(4))_(2)SO_(4)` required = 4
`rArr` Mass of `(NH_(4))_(2)SO_(4)` required `= 4 xx 132 = 528 g`