Question

3g of h2 reacts with 29g of o2 calculate the amount of water produced and amount of unreacted reactant

Answer 1

Answer:

Explanation:

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

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