Question

3 g of H2 reacts with 29 g O2 to yield H2O.
Find the limiting reagent. Calculate maximum amount of water that can be formed. Calculate the amount of one of the reactants which remains unreatced

Answer 1

HEY USER.
HERE IS THE ANSWER---
The balanced equation for the above reaction is as follows:

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.

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

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

HOPE THIS HELPS U.

Answer 2

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

Hope This Helps You