One mole of nitrogen gas on reaction with 3.0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

When $22.4 \ L$ of $H_{2(g)}$ is mixed with $11.2 \ L$ of $Cl_{2(g)}$,each at $S.T.P.$,the moles of $HCl_{(g)}$ formed is equal to

Calculate the amount of carbon dioxide produced when:
$(i)$ $1$ mole of carbon is burnt in air.
$(ii)$ $1$ mole of carbon is burnt in $16 \ g$ of dioxygen.
$(iii)$ $2$ moles of carbon are burnt in $16 \ g$ of dioxygen.

$1.0 \ g$ of $Mg$ is burnt with $0.28 \ g$ of $O_2$ in a closed vessel. Which reactant is left in excess and how much?

What are $x$ and $y$ in the following reaction?
$x Pb_3O_4 \rightarrow y PbO + O_2$

Calculate the number of moles of oxygen required to react with $6.8 \, g$ of ammonia in the following reaction:
$NH_3 + O_2 \to NO + H_2O$

Vedclass Test Series

Exam Paper Generator

Online Exam Module