Question

Calculate the number of oxygen molecules 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. $1.8\times {10}^{23}$
• C. $1.8\times {10}^{21}$
• D. $5.4\times {10}^{21}$

Answer 1

The correct option is B $1.8\times {10}^{23}$

Mass of nitrogen given $=7g$
Mass of niitrogen reacted $=80\%of7g=5.6g$

$N_2+\frac{3}{2}{O}_2\to N_2{O}_3$

$28g{N}_2$ reacts with $48g{O}_2$ to give $N_2{O}_3$, so
$5.6g$ of $N_2$ will react with $\frac{48}{28}\times 5.6$ g of $O_2$

$32g$ of $O_2$ has $6.022\times {10}^{23}$ molecules,
$(\frac{48}{28}\times 5.6)$ g of $O_2$ has

$\frac{6.022\times {10}^{23}}{32}\times \frac{48}{28}\times 5.6=1.8\times {10}^{23}molecules$

Hence the answer is B