What quantity of ammonium sulphate is necessary for the production of NH3 gas sufficient to neutralize a solution?

What quantity of ammonium sulphate is necessary for the production of ammonia gas sufficient to neutralize a solution containing $$292$$ g of HCl?
($$HCl= 36.5, (NH_{4})_{2}SO_{4}=132, NH_{3}=17$$)

A
272 g
B
408 g
C
528 g
D
1056 g

Eq. of $$(NH_{4})_{2}SO_{4} = $$Eq. of $$HCl$$.
Let, the amount of $$(NH_{4})_{2}SO_{4}$$ needed is $$x$$ g.

Then, $$\displaystyle \frac{x}{132/2}=\frac{292}{36.5}$$.Hence, $$\mathrm{x}=528$$ g.

What quantity of (NH4)SO2 is necessary for the production of NH3 gas sufficient to neutralize a solution containing 292 g of HCl

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272 g403 g528 g1056 g

Answer : C

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