Question

2) 50.0 kg of N2 (g) and 10.0 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 NH, in this situation.​

Answer 1

Answer:

50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g).

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 103 g/ 1 kg x 2 = 4.96X 103 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 103 moles.

Here we have 4.96X 103 mol of hydrogen. Hence Hydrogen is the limiting reagent.

Let us calculate the amount ammonia formed by reacting 4.96X103 moles Hydrogen

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

4.96 x103 moles Hydrogen -----?

= 4.96 x103 X ⅔

= 3.30 x 103 moles of NH3.