3 moles of nitrogen and 5 moles of hydrogen react to form ammonia. Use the balanced equation below to determine which is the limiting reactant

Question

1/ 3 moles of nitrogen and 5 moles of hydrogen react to form ammonia. Use the balanced equation below to determine which is the limiting reactant.

                               N₂ + 3H₂ --> 2NH₃

2/ Nitrogen gas and hydrogen gas are combined in a reaction to produce ammonia. If 6.0 moles of N₂ and 6.0 moles of H₂ are present in the mixture, which is the limiting reactant?

3/ For the following reaction, identify the limiting reactant and calculate the grams of product produced when 40.0 grams of the first reactant and 25.0 grams of the second reactant are used:

              2SO₂ + O₂ --> 2SO₃   

              You will have two answers for this question –

1. Limiting reactant
2. amount of SO₃ produced (grams)

Answer 1

1)

3 mol N₂ x (2 NH₃ / 1 N₂) = 6.0 mol NH₃

5 mol H₂ x (2 NH₃ / 3 H₂) = 3.3 mol NH₃

Answer: hydrogen is the limiting reactant.

2)

6 mol N₂ x (2 NH₃ / 1 N₂) = 12.0 mol NH₃

6 mol H₂ x (2 NH₃ / 3 H₂) = 4.0 mol NH₃

Answer: hydrogen is the limiting reagent.

3)

2SO₂ + O₂ >> 2SO₃   

40.0g SO₂ x (mol / 64.066g) = 0.624 mol SO₂ x (2 mol SO₃ / 2 mol SO₂) = 0.624 mol SO₃

25.0g O₂ x (mol / 31.988g) = 0.782 mol O₂ x (2 mol SO₃ / 1 mol O₂) = 1.56 mol SO

0.624 mol SO₃ x (80.06g / mol) = 50.0g SO₃

Answer:

1. Limiting reactant: SO₂
2. Product: 50.0g SO₃

Answer 2

this video will help you get the right answer

https://youtu.be/45mKzpJ5iIg