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 $O_2$.

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

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

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 $\frac{0.25}{5}\times 4=0.20$ 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 $O_2$.