Question

56 kg of nitrogen and 10kg of Hydrogen are mixed to produce ammonia calculate the no.of moles ammonia gas formed

Answer 1

Let us write the balanced equation

N2 + 3H2 → 2NH3

Now calculate the number of moles

Number of moles of N2 = 50 kg of N2 = 50 X 10^3 g/1 kg x 28g = 17.86 x 10^2 mole

Number of moles of H2 = 10 kg of N2 = 10 X 10^3 g/ 1 kg x 2    = 4.96X 10^3 mol

According to the above equation 1 mole of N2 reacts with  3 moles H2.

That is 17.86 x 10^2 mole of N2 reacts with ------moles of H2

= 3/1 X 17.86 x 10^2 = 5.36 x 10^3 moles.

Here we have 4.96X 10^3 mol of hydrogen. Hence Hydrogen is the limiting reagent.

Let us calculate the amount ammonia formed by reacting 4.96X10^3 moles Hydrogen

3 moles of hydrogen -------2 moles of NH3

4.96 x10^3 moles Hydrogen  -----?

= 4.96 x10^3 X ⅔

= 3.30 x 10^3 moles of NH3

Answer 2

Answer:

     N2(g) + 3H2(g) ⇌ 2NH3(g)

Since, 28 kg N2 reacts with 6 kg of H2

Therefore, 56 kg N2 reacts with 628628 x 56 = 12 kg of H2

But we have only 10 kg of H2, therefore, H2 is limiting reactant.

Also, 6 kg of H2 will give 2 moles of NH3

 Hence, 10 kg of H2 will give 2626 × 10  = 3.33