Question

39g of H₂ react with 29g of O₂ to give H₂O. Which is the limmiting reagent? Calculate the maximum amount of H₂O that can be formed. Calculate the amount of one of the reactants that remains unreacted.

Answer 1

3g of hydrogen will be the limiting reagent.
2H2+O2 ------->2H2O
From the above equation it is clear that 2 mole H2 react with 1 mole O2

Molar mass of H2 = 2g

Molar mass of O2= 32 g

​This implies,

4 g H2 react with 32 g O2

3 g H2 reacts with = (32/4) x 3g of O2 gas
= 24 g
As the given amount of O2 is more than required therefore O2 is excess reagent and H2 is limiting reagent.

B.
2 mole of hydrogen gas reacts to form 2 mole of water molecule therefore,
4 g of H2 produces = 36 g of water

So the amount of H2O produced by 3 g H2 = (36/4) x 3
= 27 g
Hence, 27 g of water will be produced during the recation

C.
As, 24 g of oxygen has been utilised during the recation and 29 g of oxygen was supplied therefore amount of oxygen gas left is (29-24) = 5g