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If you have $50$ grams of ${N_2}$ how much $N{H_3}$ can you make?

Last updated date: 10th Aug 2024
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Hint: To solve this question, we can use the concept of stoichiometry. Stoichiometry can be defined as a branch of chemistry that involves using the relationships between reactants or products in a chemical reaction to obtain the desired quantitative data. Stoichiometric coefficient is the number which is written in front of the atoms, ions, or molecules in a chemical reaction in order to balance the number of each element on both sides of the equation.

As we know the process of producing ammonia by combining nitrogen from the air and hydrogen from natural gas like methane is called the Haber process. This reaction is reversible and the production of ammonia is said to be exothermic.
The reaction can be written as
${N_2}(g) +3{H_2}(g) \xrightarrow {Fe\, catalyst,450^o C,200 atm} 2N{H_3}$
From this we can deduce the mole ratio between ${N_2}$ and $N{H_3}$
By mole ratio it means a conversion factor that relates the amount of a substance in moles, which in this case is $1:2$ because from the equation we know that for every mole of nitrogen gas used two moles of ammonia is produced.
So as in the question, $50$ grams of nitrogen gas is provided from which we will first find out the number of moles of nitrogen.
Molar mass of nitrogen is $28g/mol$
Hence,
$Number{\text{ of moles of }}{{\text{N}}_2} = \dfrac{{50g}}{{28g/mol}} = 1.79mol$
Therefore the number of moles of ammonia will be $2 \times 1.79 = 3.58moles$since two moles of ammonia is produced from one mole of nitrogen.
Molar mass of ammonia is $17.031g/mol$
Hence,
$amount{\text{ }}of{\text{ }}ammonia{\text{ }}that{\text{ }}is{\text{ }}produced = 3.58 \times \dfrac{{17.031g}}{{mol}} = 61g$
Therefore we can say $61g$ of ammonia is produced from $50$ grams of nitrogen.

Note:Stoichiometry is based on the law of conservation of mass where the total mass of the reactants is equal to the total mass of the products, which means that the relations among quantities of reactants and products form a ratio of positive integers. Hence, we can say that, if the amounts of the separate reactants are known, then the amount of the product can be calculated.