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.

Step-by-step 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}=(21.6g)/(30.01g/mol)=0.720moles`

b) moles of
`O_2`

`\text{Number of moles}=(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=
`(1)/(2)* 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