Question

0kg of n2(g) and 10.0kg of h2(g) are mixed to produce nh3(g) .Identify the limiting reagent and find the amount of nh3(g) formed

Answer 1

Answer:

Explanation:

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 4.96X103 moles Hydrogen

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

4.96 x10^3 moles Hydrogen -----?

= 4.96 x10^3 X 2/3

= 3.30 x 10^3 moles of NH3