To calculate the amount of NH3 formed, we need to use the balanced chemical equation for the reaction between N2 and H2 to produce NH3. The balanced chemical equation is:

N2 + 3H2 → 2NH3

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 formed can be calculated as follows:

- Calculate the number of moles of N2 in 50.0 kg:

Molar mass of N2 = 28 g/mol
Number of moles of N2 = mass/molar mass = 50,000 g/28 g/mol = 1785.71 mol

- Calculate the number of moles of H2 in 10 kg:

Molar mass of H2 = 2 g/mol
Number of moles of H2 = mass/molar mass = 10,000 g/2 g/mol = 5000 mol

- Identify the limiting reagent:

To identify the limiting reagent, we need to calculate the number of moles of NH3 that can be produced from each of the reactants. The reactant that produces the least amount of NH3 is the limiting reagent.

From the balanced chemical equation, we can see that 1 mole of N2 produces 2 moles of NH3, while 1 mole of H2 produces only 2/3 mole of NH3. Therefore, the limiting reagent is H2, as it produces the least amount of NH3.

- Calculate the amount of NH3 formed from the limiting reagent:

Number of moles of NH3 produced from 5000 mol of H2 = 5000 mol × 2/3 mol NH3/mol H2 = 3333.33 mol

- Convert the number of moles of NH3 to mass:

Molar mass of NH3 = 17 g/mol
Mass of NH3 formed = number of moles × molar mass = 3333.33 mol × 17 g/mol = 56,666.67 g = 56.67 kg

Therefore, the amount of NH3 formed is 56.67 kg.



The limiting reagent in a chemical reaction is the reactant that is completely consumed in the reaction, thereby limiting the amount of product that can be formed. In this case, we have 50.0 kg of N2 and 10 kg of H2, and we want to produce NH3. 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 reactant that produces the least amount of NH3 is the limiting reagent.

To identify the limiting reagent, we need to calculate the number of moles of NH3 that can be produced from each of the reactants. From the balanced chemical equation, we can see that 1 mole of N2 produces 2 moles of NH3, while 1 mole of H2 produces only 2/3 mole of NH3. Therefore, the limiting reagent is H2, as it produces the least amount of NH3

Mass of N2 taken =50 kg =50, 000 gm
n=gm/mm=50000/14(2) =1785.7mol

Reaction= N2+3H2--->2NH3

mass of H2 taken=10kg=10, 000gm
n=gm/mm=10000/2(1) =5000mol

clearly Hydrogen us the limiting reagent as it is present in limited quantity (5000mol)

from the balanced equation :
3 mol of H2= 2mol of NH3
or 3 mole of H2=34 g NH3
1 mol H2=34/3 of NH3
so, 5000 mol of H2 will produce
34/3×5000 kg of NH3
=56, 666.66kg
=56.66 g
of NH3