Question

The industrial production of ammonia (NH3) involves reacting nitrogen gas with hydrogen gas. If 18.2 kg of ammonia is produced from a reaction mixture that contains 6.0 kg of hydrogen and 30.0 kg of nitrogen, what is the percent yield of the reaction?

Answer 1

Answer:

The percent yield of the reaction is 53.5 %

Explanation:

This is the reaction:

3H₂ (g) +  N₂ (g) → 2NH₃(g)

Let's convert the mass we have in moles:

Mass / Molar mass: Moles

Notice that molar mass is in g/m, so we have to convert the mass of reactants from kg to g.

18.2 kg = 18200 g

6 kg = 6000 g

30 kg = 30000 g

6000 g/ 2g/m = 3000 moles H₂

30000 g / 28 g/m = 1071.4 moles N₂

18200 g / 17 g/m = 1070.5 moles NH₃

Ratio of reactants is 3:1 . It's easy to see, that limiting reactant is the hydrogen.

1 mol of N₂ needs 3 moles of H₂

1071.4 mol of N₂ need ____ 1071.4  .3 = 3214.2 moles

I only have 3000 moles of H₂

Let's go to the rule of three

So 3 moles of H₂ __ are needed to make 2 moles of NH₃

3000 moles of H₂ _ are needed to make ( 3000 .2) /3 = 2000 moles

2000 moles are produced if the reaction was at 100 % yield, but we only produced 1070.5 moles of amonia.

So the last rule of three will be:

2000 moles ____ 100 % yield

1070.5 moles ____ (1070.5 . 100) /2000 = 53.5%