Question

Calculate the number of oxygen atoms required to combine with $7g$ of $N_{2}$ to form $N_{2}O_{3}$ when $80\%$ of $N_{2}$ is converted to $N_{2}O_{3}$ .

• A. $2.3\times 10^{23}$
• B. $3.6\times 10^{23}$
• C. $1.8\times 10^{23}$
• D. $5.4\times 10^{21}$

Answer 1

Answer: (B)

for formation of $N_{2}O_{3}$
$2N_{2}+3O_{2} \rightarrow 2N_{2}O_{3}$
$2$ moles of $N_{2}$ combines with $3mole$ of $O_{2}$
Given, $7gm$ $N_{2}$ $\left(\right.80\%$ only reacts $\left.$ $i.e.5.6$ $gm$ $of$ $N_{2}$
Mole of $N_{2}=\frac{5.6}{28}=0.2$
Hence $\frac{2}{0.2}=\frac{3}{x m o l e o f O_{2}}$
$x=0.3$ moles
Number of atoms $=0.3\times 2\times 6.023\times 10^{23}$
$=3.61\times 10^{23}$