Question

If 3.0 g of NH3 reacts with 5.0 g of HCl, what is the limiting reactant?

Answer 1

Answer:HCl

Explanation:

$\text{number of moles}=\frac{\text{given mass}}{\text{molar mass}}$

For $NH_{3},$

$\text{given mass}=3g$

$\text{molar mass}=14+3=17g$

$n_{NH_{3}}=\frac{3}{17}=0.176$

For $HCl,$

$\text{given mass}=5g$

$\text{molar mass}=1+35.5=36.5g$

$n_{HCl}=\frac{5}{36.5}=0.136$

The reaction between $NH_{3}$ and $HCl$ is

$NH_{3}+HCl$→$NH_{4}Cl$

So,one mole of $NH_{3}$ requires one mole of $HCl$

$0.176$ moles of $NH_{3}$ requires $0.176$ moles of $HCl$

But there are only $0.136$ moles of $HCl$ available.

$HCl$ will be consumed first.

So,$HCl$ is the limiting reagent.