At NTP,40 , mL of nitrogen reacts with hydrog | vedclass

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(C) The given reaction is $N_2 + 3H_2 \rightarrow 2NH_3$.Given volume of $N_2 = 40 \, mL$ at $NTP$.First,calculate the mass of $N_2$ used:$Moles \, of

Vedclass · Accepted Answer

$0.0607$

Vedclass · Answer

(C) The given reaction is $N_2 + 3H_2 ightarrow 2NH_3$.Given volume of $N_2 = 40 \, mL$ at $NTP$.First,calculate the mass of $N_2$ used:$Moles \, of \, N_2 = \frac{40 \, mL}{22400 \, mL/mol} = \frac{1}{560} \, mol$.$Mass \, of \, N_2 = Moles 	imes Molar \, mass = \frac{1}{560} 	imes 28 \, g/mol = 0.05 \, g$.According to the stoichiometry of the reaction,$28 \, g$ of $N_2$ produces $34 \, g$ of $NH_3$.Therefore,the mass of $NH_3$ produced by $0.05 \, g$ of $N_2$ is:$Mass \, of \, NH_3 = \frac{34 \, g \, NH_3}{28 \, g \, N_2} 	imes 0.05 \, g \, N_2 = 0.0607 \, g$.

At $NTP$,$40 \, mL$ of nitrogen reacts with hydrogen to form ammonia. According to the reaction $N_2 + 3H_2 \rightarrow 2NH_3$,the mass of ammonia formed will be ..... (in $, g$)

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