Question

50.0kg of N2(g) and 10.0kg of H2(g) are mixed to produce NH3(g). Calculate the NH3(g) formed. Identify the limiting of NH3 in this situation.

Answer 1

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

Answer 2

Answer will be 169.92 kg.