Question

Using the equation ${N_2} + 3{H_2} \to 2N{H_3}$ , if $28g$ ${N_2}$ react, how many grams of $N{H_3}$ will be produced?

Answer 1

We need to understand and study the stoichiometry of the given balanced chemical equation. The mole concept is the key to solving problems. Mole concept is a convenient concept for expressing the amount of a given substance. In chemistry, a mole is defined as the amount of a substance which contains Avogadro’s number of particles. We will now study the mole concept of the given reaction.

Complete step by step answer:
The given equation is ${N_2} + 3{H_2} \to 2N{H_3}$ , whose mole ratio is definitely $1:3:2$ . It is clear that the nitrogen is the limiting reagent as the reaction will stop once the nitrogen is fully consumed. Hence nitrogen will be the reactant which will control the production of $N{H_3}$ . We now calculate the number of moles of both the nitrogen and the produced ammonia.
Moles of ${N_2}$ present$= 28 \times \dfrac{1}{{28g/mol}} = 1mol$
Moles of $N{H_3}$ produced = $1mole{N_2} \times 2molesN{H_3}/mole{N_2} = 2moles$ of $N{H_3}$ .
These are calculated from the fact that the number of moles of a substance can calculated by the following formula given below:
$n = \dfrac{N}{{{N_A}}}$
Where $n$ is the number of moles of the substance $N$ is the total number of entities of the particular element in the sample, and ${N_A}$ is the Avogadro’s constant whose value is $6.022 \times {10^{23}}$.
Therefore, the grams of $N{H_3}$ that can be produced are 2 moles of $N{H_3}$ multiplied by its molecular weight which is $17g/mol$ . This equals $34grams$ of $N{H_3}$ produced.

Hence, if $28g$ ${N_2}$ react, grams of $N{H_3}$ will be produced is $34$ grams.

Note: It must be noted that since there is only one mole of the limiting reagent, its constituent atoms of the product will contain a number of moles which is equal to the number of atoms. The number of moles of a molecule may not always be equal to the number of moles of its constituent elements. For example, a mole of water contains ${N_A}$ number of ${H_2}O$ molecules. However, each water molecule contains $2$ hydrogen atoms and one oxygen atom. Therefore, one mole of ${H_2}O$ contains 2 moles of hydrogen and one mole of oxygen.