What quantity of ammonium sulphate is necessa | vedclass

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(C) The chemical reaction for the production of $NH_3$ from ammonium sulphate is: $(NH_4)_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2NH_3 + 2H_2O$. The neu

Vedclass · Accepted Answer

$528$

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(C) The chemical reaction for the production of $NH_3$ from ammonium sulphate is: $(NH_4)_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2NH_3 + 2H_2O$. The neutralization reaction is: $NH_3 + HCl ightarrow NH_4Cl$. Combining these,the stoichiometry is: $(NH_4)_2SO_4 \equiv 2NH_3 \equiv 2HCl$. Given molar masses: $(NH_4)_2SO_4 = 132 \ g/mol$ and $HCl = 36.5 \ g/mol$. From the stoichiometry,$1 \ mol$ of $(NH_4)_2SO_4$ reacts with $2 \ mol$ of $HCl$. $132 \ g$ of $(NH_4)_2SO_4$ reacts with $2 	imes 36.5 \ g = 73 \ g$ of $HCl$. For $292 \ g$ of $HCl$,the amount of $(NH_4)_2SO_4$ required is: $(132 / 73) 	imes 292 = 132 	imes 4 = 528 \ g$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

What quantity of ammonium sulphate is necessary for the production of $NH_3$ gas sufficient to neutralize a solution containing $292 \ g$ of $HCl$? $(HCl=36.5, (NH_4)_2SO_4=132, NH_3=17)$

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