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