Question

3g of H₂ react with 29g of O₂ to yield H₂O, which is the limiting reactant? Calculate the maximum amount of H₂O that can be formed. Calculate the amount of the reactant which remains unreacted.

Answer 1

2H₂(g)+O₂(g)⟶2H₂O(aq)

The molar mass of H₂ = 2 g
The molar mass of O₂ = 32 g

• From the above equation, it is clear that 2 mole H₂ reacts with 1 mole O₂
• 4 g of H₂ reacts with 32 g of O₂
• 3 g of H₂ reacts with [(32/4)×3] g of O₂
• 3 g of H₂ reacts with 24 g of O₂

So, 3 g of hydrogen will be the limiting reagent. As, the given amount of O₂ is more than required, therefore O₂ is the excess reagent and H₂ is the limiting reagent.

• From the above equation, 2 moles of H₂ reacts to form 2 moles of H₂O,
• Therefore, 4 g of H₂ produces = 36 g of H₂O
• So, the amount of H₂O produced by 3 g H₂ = [(36/4) x 3] g of H₂O
  = 27 g of H₂O

Hence, 27 g of H₂Owill be produced during the reaction.
As 24 g of oxygen has been utilized during the reaction and 29 g of oxygen was supplied.

Therefore, the amount of oxygen gas left is (29-24) g = 5g