Question

What is the maximum number of grams of ammonia, nh3, which can be obtained from the reaction of 10.0 g of h2 and 80.0 g of n2? n2 + 3h2 → 2nh3?

Answer 1

Answer: The mass of ammonia produced is 28.22 g

Explanation:

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

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

• For hydrogen gas:

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol$

• For nitrogen gas:

Given mass of nitrogen gas = 80.0 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen gas}=\frac{80.0g}{28g/mol}=2.86mol$

The given chemical equation follows:

$N_2+3H_2\rightarrow 2NH_3$

By Stoichiometry of the reaction:

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas

So, 5 moles of hydrogen gas will react with = $\frac{1}{3}\times 5=1.66mol$ of nitrogen gas

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

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of hydrogen gas produces 1 mole of ammonia

So, 5 moles of hydrogen gas will produce = $\frac{1}{3}\times 5=1.66moles$ of ammonia

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 1.66 moles

Putting values in equation 1, we get:

$1.66mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(1.66mol\times 17g/mol)=28.22g$

Hence, the mass of ammonia produced is 28.22 g