Question

Calculate the amount of ammonia (NH3) formed when 2.8g of Nitrogen combines with hydrogen. (N = 14u H = 1u)​

Answer 1

Answer:

Step-by-step explanation:

N2 + H2 = NH3

After balancing  - N2 + 3H2 = 2NH3

1 Mole of nitrogen ≡ 3 mole of hydrogen

Molecular mass of nitrogen = 2 *14 = 28g

No of moles of nitrogen = 2.8 / 28 = 1/10 moles = 0.1 mole

0.1 mole of nitrogen ≡ 0.3 mole of hydrogen

no of moles of hydrogen = 0.3 = given mass / 2

mass = 0.6g

1 mole of hydrogen ≡ 1/3 mole of nitrogen

1/3 = given mass / 28

mass = 28 / 3

This tell us that nitrogen is the excess reactant and hydrogen is the limiting reagent

1 Mole of hydrogen ≡ 2 mole of NH3

3 mole of hydrogen ≡ 6 mole of NH3

No of moles = given mass / molecular mass

6 = mass / 17

mass = 6 * 17 = 102g

∴the amount of ammonia produced is 102g

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