50 kg of N2(g) and 10 kg of H2(g) are mixed to produce NH3(g) calculate the amount of NH3(g) formed.Identify the limiting reagent in the production of NH3 in this situation?

Question

Answer

Answer

N2 + 3H2 ---->2NH3

here.

28kg N reacts with 6kg of H

so,

50 kg of M react with 6/28�50kg H = 10.7kg but we have only 10 so H2 is limiting reagent..

6kg H2 gives 34kg NH3

10kg H2 will give 34/6�10=56.67kg of NH3

Answer

Calculation of Amount of NH3(g) formed

- Write the balanced chemical equation for the production of NH3(g):

N2(g) + 3H2(g) → 2NH3(g)

- Determine the moles of N2 and H2:

Moles of N2 = mass of N2 / molar mass of N2
Moles of N2 = 50 kg / 28 g/mol
Moles of N2 = 1785.71 mol

Moles of H2 = mass of H2 / molar mass of H2
Moles of H2 = 10 kg / 2 g/mol
Moles of H2 = 5000 mol

- Determine the limiting reagent:

From the balanced chemical equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the amount of NH3 that can be produced is dependent on the amount of H2 available. Thus, H2 is the limiting reagent.

- Calculate the moles of NH3 produced using the limiting reagent:

Moles of NH3 = moles of H2 / 3 x stoichiometric coefficient of NH3
Moles of NH3 = 5000 mol / 3 x 2
Moles of NH3 = 833.33 mol

- Calculate the mass of NH3 produced:

Mass of NH3 = moles of NH3 x molar mass of NH3
Mass of NH3 = 833.33 mol x 17 g/mol
Mass of NH3 = 14166.67 g or 14.17 kg

Identification of Limiting Reagent

- Definition of Limiting Reagent:

The limiting reagent is the reactant that is completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed.

- Determination of Limiting Reagent:

From the calculations above, we can see that the moles of N2 available (1785.71 mol) are much greater than the moles of H2 available (5000 mol). This means that all of the H2 will be used up in the reaction, while there will be excess N2 remaining. Therefore, H2 is the limiting reagent in the production of NH3.

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