Question

A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction goes to completion, how much of which reactant is left in excess?

Answer 1

Answer :  The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of $NH_4Cl$ = 26.5 g

Mass of $NaOH$ = 10 g

Molar mass of $NH_4Cl$ = 53.5 g/mole

Molar mass of $NaOH$ = 40 g/mole

First we have to calculate the moles of $NH_4Cl$ and $NaOH$.

$\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles$

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NH_4Cl+NaOH\rightarrow NH_4OH+NaCl$

From the balanced reaction we conclude that

As, 1 mole of $NaOH$ react with 1 mole of $NH_4Cl$

So, 0.25 moles of $NaOH$ react with 0.25 moles of $NH_4Cl$

From this we conclude that, $NH_4Cl$ is an excess reagent because the given moles are greater than the required moles and $NaOH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of $NH_4Cl$.

$\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl$

$\text{Mass of }NH_4Cl=(0.245mole)\times (53.5g/mole)=13.1075g$

Therefore, the amount of reactant left in excess is 13.1075 grams.