Question

If 20.5 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?

Answer 1

Answer: The number of moles of excess reactant $O_2$  left over will be, 0.089 moles.

Explanation : Given,

Mass of $NO$ = 20.5 g

Mass of $O_2$ = 13.8 g

Molar mass of $NO$ = 30 g/mol

Molar mass of $O_2$ = 32 g/mol

First we have to calculate the moles of $NO$ and $O_2$.

$\text{Moles of }NO=\frac{\text{Given mass }NO}{\text{Molar mass }NO}$

$\text{Moles of }NO=\frac{20.5g}{30g/mol}=0.683mol$

and,

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}$

$\text{Moles of }O_2=\frac{13.8g}{32g/mol}=0.431mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$2NO+O_2\rightarrow 2NO_2$

From the balanced reaction we conclude that

As, 2 mole of $NO$ react with 1 mole of $O_2$

So, 0.683 moles of $NO$ react with $\frac{0.683}{2}=0.342$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $NO$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of excess reactant $O_2$  left over.

Number of moles of excess reactant $O_2$  left over = Given moles - Required moles

Number of moles of excess reactant $O_2$  left over = 0.431 mol - 0.342 mol

Number of moles of excess reactant $O_2$  left over = 0.089 mol

Therefore, the number of moles of excess reactant $O_2$  left over will be, 0.089 moles.