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How many grams of NH3 form when 22g H2 react completely?

Sagot :

The Answer to the question is 124.66 grams approximately.

It is given that H₂ will react completely

Molar mass of H₂ = 2 × 1 = 2 gm/mol

Given mass of H₂ = 22 gm

Moles of H₂ reacting = given mass of H₂ / molar mass of H₂

Moles of H₂ reacting = 22 / 2 moles = 11 moles

So, 11 moles of H₂ are taking part in reaction.

The Balanced chemical reaction is as follows as:

N₂ + 3 H₂ → 2 NH₃

It is clear from the reaction that 3 moles of H₂ are making 2 mole of NH₃

3 moles of H₂ → 2 mole of NH₃

1 moles of H₂ → 2/3 mole of NH₃

11 moles of H₂ → (2/3)×11 mole of NH₃

11 moles of H₂ → 22/3 mole of NH₃

So, 11 moles of H₂ will make 22/3 mole of NH₃

22/3 mole NH₃ = mass formed NH₃ / molar mass of NH₃

mass formed NH₃ = 22/3 × molar mass of NH₃

mass formed NH₃ = 22/3 × 17

mass formed NH₃ = 22/3 × 17

mass formed NH₃ = 124.66 grams    (approximately)

124.66 grams of NH₃ will be formed.

So, approximately 124.66 grams of NH₃ is formed from 22 grams of H₂.

Learn more about Stoichometry here:

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