Question

3g H2 reacts with 29g O2 toyeid H2O. i)which is the limiting reactant ? ii)Calculate the maximum amount of H2O that can be formed. iii)Calculate the amount of one of the reactants which unreacted.

Answer 1

Answer:

Calculate maximum amount of water that can be formed. ... above equation it is clear that 2 mole H2 react with 1 mole O2 ... 3 g H2 reacts with = (32/4) x 3g of O2 gas

Answer 2

Explanation:

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 oxgen gas left is (29-24) = 5g

HOPE THIS HELPS U.