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Question: What is limiting reagent? \(50kg{\text{ }}{N_2}\) and \(10kg{\text{ }}{H_2}\) are mixed to produce \...

What is limiting reagent? 50kg N250kg{\text{ }}{N_2} and 10kg H210kg{\text{ }}{H_2} are mixed to produce NH3N{H_3}. Calculate the ammonia gas formed. Also, identify the limiting reagent.

Explanation

Solution

The limiting reagent needs to be identified in a reaction to calculate the percentage yield and the theoretical yield is defined as the amount of product obtained when the limiting reagent reacts completely.

Complete step by step answer:

Limiting reagent is defined as the reactant that is completely used and it determines the amount of product formed. The reaction cannot continue once the limiting reagent is completely used
To calculate the amount of ammonia produce when 50kg N250kg{\text{ }}{N_2}and 10kg H210kg{\text{ }}{H_2} are mixed we have to write a balanced chemical equation as.
N2+3H22NH3{N_2} + 3{H_2} \to 2N{H_3}
As we can see one mole of nitrogen reacts with three mole of hydrogen gas. The molecular mass of nitrogen gas is 28 g/mol28{\text{ }}g/mol and the molecular mass of hydrogen is 3×2=6kg/mol3 \times 2 = 6kg/mol.
Since we know 1 kg =1000g1{\text{ }}kg{\text{ }} = 1000g, so after converting the masses to gram (g), we get molecular mass of nitrogen gas and hydrogen gas as 0.028kg/mol and 0.006kg/mol0.028kg/mol{\text{ }} and {\text{ }}0.006kg/mol respectively. According to the balanced chemical equation:
0.028kg of nitrogen react = 0.006kg of hydrogen
\Rightarrow1kg of nitrogen react =0.0060.028kg\dfrac{{0.006}}{{0.028}}kg of hydrogen
\Rightarrow50kg of nitrogen react = 0.0060.028×50=10.71kg\dfrac{{0.006}}{{0.028}} \times 50 = 10.71kg of hydrogen.
Hence the amount of hydrogen is given as 10 kg which is lesser as compared to the amount required to react. Therefore the hydrogen is a limiting reagent and completely consumed during the reaction and the amount of ammonia formed depends upon the amount of hydrogen.
Again according to the chemical equation:
3 mole of hydrogen gas i.e. 0.006 kg of hydrogen produced = 2 mole of ammonia i.e. 34g or 0.034kg34g{\text{ or }}0.034kg of ammonia.
Hence we can write 0.006kg hydrogen produced= 0.034 kg of ammonia.
1 kg of hydrogen produced= 0.0340.006kg\dfrac{{0.034}}{{0.006}}kg of ammonia.
10kg of hydrogen produced = 0.0340.006×10=56.67kg\dfrac{{0.034}}{{0.006}} \times 10 = 56.67kg of ammonia.
Hence the amount of ammonia produced when 50kg N250kg{\text{ }}{N_2}and 10kg H210kg{\text{ }}{H_2} are mixed is 56.67kg.

Note:
After the limiting reagent is consumed, there is nothing left from the other reactant to react with and hence, they will not be completely consumed and are known as excess reagent.