Question

Calculate the number of oxygen atoms required to combine with $7.0$ g of $N_2$ to form $N_2{O}_3$ if $80\%$ of $N_2$ is converted into products.
$N_2+\frac{3}{2}{O}_2\to N_2{O}_3$

• A. $3.24\times {10}^{23}$
• B. $3.6\times {10}^{23}$
• C. $18\times {10}^{23}$
• D. $6.02\times {10}^{23}$

Answer 1

Answer: (B)

$3.6\times {10}^{23}$

$N_2+3/2O_2\to N_2{O}_3$
As $1$ mol ( $28$ g) combine with $1.5$ mol ($32\times 3/2=48$ g) of $O_2$.
Then $\frac{7}{28}\times \left(80/100\right)$ moles of $N_2$ will combine with $1.5\times \frac{7}{28}\times \left(80/100\right)=0.3$ mol of $O_2$
$0.3$ moles of $O_2=6.022\times {10}^{23}\times 0.3=1.8\times {10}^{23}$ molecules of $O_2$
$1.8\times {10}^{23}$ molecules of $O_2=3.6\times {10}^{23}$ O-atoms.