For the formation of 3.40 g of ammonia gas,w | vedclass

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(C) The balanced chemical equation for the formation of ammonia is:$N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$According to the stoichiometry:$1 \ mol$

Vedclass · Accepted Answer

$6.72 \ L$ and $2.24 \ L$

Vedclass · Answer

(C) The balanced chemical equation for the formation of ammonia is:$N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$According to the stoichiometry:$1 \ mol$ of $N_2$ ($22.4 \ L$ at $NTP$) reacts with $3 \ mol$ of $H_2$ ($3 	imes 22.4 \ L = 67.2 \ L$ at $NTP$) to produce $2 \ mol$ of $NH_3$ ($2 	imes 22.4 \ L = 44.8 \ L$ at $NTP$).The molar mass of $NH_3$ is $17 \ g/mol$. Therefore,$2 \ mol$ of $NH_3$ corresponds to $2 	imes 17 = 34 \ g$.For $34 \ g$ of $NH_3$,we require $67.2 \ L$ of $H_2$ and $22.4 \ L$ of $N_2$.For $3.40 \ g$ of $NH_3$ (which is $0.1$ times $34 \ g$),the required volumes are:Volume of $H_2 = 0.1 	imes 67.2 \ L = 6.72 \ L$Volume of $N_2 = 0.1 	imes 22.4 \ L = 2.24 \ L$Thus,the required volumes are $6.72 \ L$ of $H_2$ and $2.24 \ L$ of $N_2$.

For the formation of $3.40 \ g$ of ammonia gas,what volumes of hydrogen gas and nitrogen gas,respectively,are required at $NTP$ conditions?

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