Question

For the balanced reaction given below, what volume of $N{O}_2$(g) is produced from the combustion of 68 g of $N{H}_3$(g), assuming the reaction takes place at standard temperature and pressure?
$4N{H}_3\left(g\right)+7O_2\left(g\right)\to 4N{O}_2\left(g\right)+6H_{2}O\left(l\right)$

• A. 22.4 L
• B. 44.8 L
• C. 67.2 L
• D. 89.6 L

Answer 1

The correct option is D 89.6 L

For reaction,
$4N{H}_3\left(g\right)+7O_2\left(g\right)\to 4N{O}_2\left(g\right)+6H_{2}O\left(l\right)$

4 mol of $N{H}_3$ produce 4 mol of $N{O}_2$
So, 4$\times$17 g $N{H}_3$ produces 4$\times$46 g of $N{O}_2$
68 g $N{H}_3$ produces 184 g of $N{O}_2$
$\because$ volume occupied by 1 mole of gas at STP = 22.4 L
$\therefore$ volume of $N{O}_2$ produced =$\frac{184\times 22.4}{46}=89.6$ L