Question

3 g of H₂ reacts with 29 g of O₂ to yield water. Which is the limiting reactant?
(A) H₂
(B) O₂
(C) H₂O
(D) none of there

Answer 1

HEYA MATE,

HERE IS UR ANSWER.

Thanks for asking this question.

First write the chemical equation.

2H₂ + O₂ -----> 2H₂O

For 2H₂ - 2 x 2 = 4 g
O₂ - 32 g
2H₂O - 2 x 18 = 36 g

Now, we have to determine the limiting reagent.

4 g of H₂ reacts with 32 g of O₂
1 g of H₂ reacts with 32/4 g of O₂
3 g of H₂ reacts with 32/4 x 3 = 24 g of O₂

But according to the question, 29 g of O₂ is present.

So, the limiting reactant is hydrogen.

Now, 4 g of H₂ forms 36 g of H₂O

1 g of H₂ forms 36/4 g of H₂O.
3 g of H₂ forms 36/4 x 3 = 27 g of H₂O

Maximum amount of water that can be formed is 27 g.

For, amount of oxygen left of unreacted,
Only 24 g of oxygen will react.

But 29 g is the given amount.
Amount of oxygen unreacted = 29 - 24 = 5 g

I HOPE IT HELPS U.