Question

50 kg of an and 10 kg of h2 are mixed together to produce ammonia identify the limiting reagent also

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 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