Question

6g H2 reacts with 24g O2 to yield H2O. a) Which is the limiting reagent? b) Calculate the maximum amount of H2O that can be formed? c) Calculate the amount of one of the reactants which remains unreacted?​

Answer 1

Answer:

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..

Mark me as brainliest...

Answer 2

Balanced equation of the reaction is

2H2 + O2 = 2H2O

From the above equation, 2 mole H2 react with 1 mole O2

• Molar mass of H2 = 2g
• Molar mass of O2= 32 g

» 4 g H2 react with 32 g O2

3 g H2 reacts with = (32/4) x 3g of O2 gas = 24 g

• 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 = 27 g

• Hence, 27 g of water will be produced

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