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

Question

Dr. Abubacker · Accepted Answer

(c)
$\left( NH_{4} ight)_{2}SO_{4} \longrightarrow 2NH_{3} + H_{2}O + SO_{3}$
$3NH_{3} + 2HCl \longrightarrow 2NH_{4}Cl$
$\left( NH_{4} ight)_{2}SO_{4} \equiv 2NH_{3} \equiv 2HCl$
132 g 73g
73g$HCl \equiv 132g\left( NH_{4} ight)_{2}SO_{4}$
292 $g\ HCl \equiv \frac{132 	imes 292}{73}g\left( NH_{4} ight)_{2}SO_{4}$
$= 528\ g\left( NH_{4} ight)_{2}SO_{4}$

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

Explanation

Reviewed by Dr. Abubacker

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