Question

What is the limiting reagent when 49.84 g of nitrogen react with 10.7 g of hydrogen according to this balanced equation: n2(g) 3h2(g) → 2nh3(g) ?

Answer 1

Moles of nitrogen = 49.84 / 28 = 1.78
Moles of hydrogen = 10.7 / 2 = 5.35

Molar ratio of nitrogen to hydrogen = 1 : 3
Hydrogen required = 1.78 x 3 = 5.34

Therefore, hydrogen is in excess and nitrogen is the limiting reagent. 

Answer 2

Answer: The limiting reagent for the given chemical reaction is hydrogen.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   .....(1)

• For Nitrogen:

Given mass of nitrogen = 49.84 g

Molar mass of nitrogen = 28 g/mol

Putting values in above equation, we get:

$\text{Moles of nitrogen}=\frac{49.84g}{28.0134g/mol}=1.78mol$

• For Hydrogen:

Given mass of hydrogen = 10.7 g

Molar mass of hydrogen = 2 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen}=\frac{10.7g}{2.01g/mol}=5.32mol$

For the given chemical reaction:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

By Stoichiometry of the reaction:

3 moles of hydrogen reacts with 1 mole of nitrogen.

So, 5.32 moles of hydrogen reacts with = $\frac{1}{3}\times 5.32=1.77moles$ of nitrogen.

As, the given amount of nitrogen is more than the required amount, so it is considered as an excess reagent.

Thus, hydrogen is considered as an limiting reagent because it limits the formation of products.

Hence, the limiting reagent for the given chemical reaction is hydrogen.