Question

If 21.6 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over? 2 NO (g) + O₂ (g) → 2 NO₂ (g)

Answer 1

Answer: 0.071 moles of excess reactant $(O_2)$ will be left over.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $NO$

$\text{Number of moles}=\frac{21.6g}{30.01g/mol}=0.720moles$

b) moles of $O_2$

$\text{Number of moles}=\frac{13.8g}{32g/mol}=0.431moles$

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

According to stoichiometry :

2 moles of $NO$ require 1 mole of $O_2$

Thus 0.720 moles of  $NO$ require=$\frac{1}{2}\times 0.720=0.360moles$ of $O_2$

Thus $NO$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess regaent.

moles of $O_2$ left = (0.431-0.360) = 0.071 moles