Nitric oxide, $N O$ , reacts with oxygen gas to form nitrogen dioxide, a dark brown gas.
What mass of nitrogen dioxide could be produced from $20.3 g$ of nitric oxide and $5.89 g$ of oxygen gas?

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Solution

$2 N O + O_{2} ⟶ 2 N O_{2}$
Number of moles of $N O$ in $20.3 g = \frac{20.3 g}{30} = 0.67$ moles
Number of moles of $O_{2}$ in $5.6 g = \frac{5.6}{32} = 0.175$ moles
As $1$ mole of $O_{2}$ is reacting with $2$ moles of $N O$
$0.175$ moles of $O_{2}$ is reacting with $2 \times 0.175$ moles of $N O = 0.35$ moles of $N O$ .
Thus $O_{2}$ is limiting agent.
Moles $2 N O + O_{2} ⟶ 2 N O_{2}$
at $t = 0$ $0.67$ $0.175$ -
After reaction $0.67 - 0.35$ - $0.35$
Thus moles of $N O_{2}$ produce= $0.35$ as $1$ mole of $O_{2}$ give $2$ moles of $N O_{2}$ .
Mass of $N O_{2}$ produced= n $\times$ Molar mass
$= 0.35 \times 46$
$\Rightarrow m_{N O_{2}} = 16.1 g$
Thus $16.1 g$ of $N O_{2}$ is produced.

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