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2 years ago

How many molecules of O₂ will be required to produce 28.8 g of water? 2H₂ + O₂ ---> 2H₂O

Chemistry

1 answer:

juin [17]2 years ago

4 0

To produce 28.8 g of water 4.81 x 10²³ molecules of O₂ will be required

<h3>What is limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

According to given equation;

2H₂ + O₂ ---> 2H₂O

If 36 g of H₂O (2 mole) is produced by 1 mole O₂ i.e, 32 gram.

Therefore, 28.8 g of H₂O is produced by ;

32/36 x 28.8 = 25.6 g of O₂

Number of molecules of O₂ = given weight / molecular weight x 6.023 x 10²³

= 25.6 / 32 x 6.023 x 10²³

= 4.81 x 10²³ molecules of O₂

Therefore,4.81 x 10²³ molecules of O₂ will be required to produce 28.8 g of water.

Learn more about Limiting reagent here ;

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