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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.


Solution

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


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