Question

What is the limiting reactant if 4.0 g of Nh3 react with 8.0 g of oxygen?

Answer 1

NH3-The limiting reactant is the reactant that get completely used up in a reaction

Answer 2

Answer : The limiting reagent is
`O_2`.

Solution : Given,

Mass of
`NH_3` = 4.0 g

Mass of
`O_2` = 8.0 g

Molar mass of
`NH_3` = 17 g/mole

Molar mass of
`O_2` = 32 g/mole

First we have to calculate the moles of
`NH_3` and
`O_2`.

`\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}=(4.0g)/(17g/mole)=0.24moles`

`\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(8.0g)/(32g/mole)=0.25moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`4NH_3+5O_2\rightarrow 4NO+6H_2O`

From the balanced reaction we conclude that

As, 5 mole of
`O_2` react with 4 mole of
`NH_3`

So, 0.25 moles of
`O_2` react with
`(0.25)/(5)* 4=0.20` moles of
`NH_3`

From this we conclude that,
`NH_3` is an excess reagent because the given moles are greater than the required moles and
`O_2` is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is
`O_2`.