One mole of nitrogen gas on reaction with 3.0 | vedclass

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Question

(B) The balanced chemical equation for the reaction is: $N_2 + 3H_2 ightleftharpoons 2NH_3$. Given: $1$ mole of $N_2$ and $3.01 	imes 10^{23}$ mole

Vedclass · Accepted Answer

$2.0 	imes 10^{23}$ molecules of ammonia

Vedclass · Answer

(B) The balanced chemical equation for the reaction is: $N_2 + 3H_2 ightleftharpoons 2NH_3$. Given: $1$ mole of $N_2$ and $3.01 	imes 10^{23}$ molecules of $H_2$. Since $1$ mole of $H_2$ contains $6.022 	imes 10^{23}$ molecules,the number of moles of $H_2$ is $\frac{3.01 	imes 10^{23}}{6.022 	imes 10^{23}} = 0.5$ mole. According to the stoichiometry,$3$ moles of $H_2$ produce $2$ moles of $NH_3$. Therefore,$0.5$ mole of $H_2$ will produce $\frac{2}{3} 	imes 0.5 = \frac{1}{3}$ mole of $NH_3$. Number of molecules of $NH_3 = \frac{1}{3} 	imes 6.022 	imes 10^{23} \approx 2.0 	imes 10^{23}$ molecules. Thus,the correct option is $B$.

One mole of nitrogen gas on reaction with $3.01 \times 10^{23}$ molecules of hydrogen gas produces

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