Question

Calculate the number of oxygen atoms required to combine with 7 g of $N_2$ to form $N_2{O}_3$ when $80\%$ of $N_2$ is converted to $N_2{O}_3$.

• A. $7.2\times {10}^{23}$
• B. $3.6\times {10}^{23}$
• C. $1.8\times {10}^{21}$
• D. $5.4\times {10}^{21}$

Answer 1

The correct option is B $3.6\times {10}^{23}$

Mass of nitrogen given = $7g$
Mass of nitrogen reacted $=80\%$ of $7g=5.6g$
Balanced chemical equation
$N_2+\frac{3}{2}{O}_2\to N_2{O}_3$

$28g{N}_2\text{reacts with}48g{O}_2\text{to give}{N}_2{O}_3$
By unitary method,$5.6g\text{of}{N}_2\text{will react with}\frac{48}{28}\times 5.6=9.6g\text{of}{O}_2$ to give $N_2{O}_3$
$16g{O}_2\text{has}6.023\times {10}^{23}\text{atoms}$
By unitary method, $9.6gof{O}_2\text{has}\frac{6.023\times {10}^{23}}{16}\times 9.6=3.61\times {10}^{23}\text{atoms of}O$
Hence the answer is B