Question

Ammonia, NH3(g), can be prepared by the following reaction:

2NO(g) + 5 H2(g)----> NH3(g) + 2 H2O(g)

What is the limiting reagent of the reaction and the theoretical yield of NH3(g) in grams if 26.5 g NO(g) and 1.77 g of H2(g) were added to a flask?

A, NO(g) and 7.51 g NH2(g)

B. NO(g) and 15.0 g NH3(g)

C. H2(g) and 15.0 g NH3(g)

D. H2(g) and 5.96 g NH3(g)

*Please explain how you got your answer. Much appreciated :)

Answer 1

2NO(g) + 5 H2(g) ----> 2NH3(g) + 2 H2O(g)

Mass of NO = 26.5 g

Molar mass of NO = 30.01 g/mole

Moles of NO = 26.5 / 30.01

= 0.88 moles

Mass of H2 = 1.77 g

Molar mass of H2 = 2.01 g/mole

Moles of NO = 1.77 / 2.01

= 0.88 moles

As 2 moles NO reacts with 5 moles H2.

So, 0.88 moles No will react with = 5/2 * 0.88

= 2.2 moles of H2.

But we have only 0.88 moles of H2.

So, H2 is limiting reagent.

Moles of NH3 produced = 2/5 * 0.88

= 0.352 moles

Molar mass of NH3 = 17.03 g/mole

Mass of NH3 produced = 17.03 * 0.352

= 5.9 g

So, theoretical yield of NH3 = 5.9 g

Hence,

option D. H2(g) and 5.96 g NH3(g) is correct.