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 $N_2{O}_3$

$2N_2+3O_2\to 2N_2{O}_3$

$2$ moles of $N_2$ combines with $3mole$ of $O_2$

Given, $7gm$ $N_2$ $(80\%$ only reacts) $i.e.5.6$ $gm$ $of$ $N_2$

Mole of $N_2=\frac{5.6}{28}=0.2$

Hence $\frac{2}{0.2}=\frac{3}{xmoleof{O}_2}$

$x=0.3$ moles

Number of atoms $=0.3\times 2\times 6.023\times 10^{23}$

$=3.61\times 10^{23}$