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=(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=(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
`(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.