Final answer:
When 10.54 g of H₂ reacts with 95.10 g of O₂, 94.04 g of water (H₂O) will be formed, as hydrogen is the limiting reactant in this reaction.
Step-by-step explanation:
The question is asking for the mass of water (H₂O) produced from the reaction of hydrogen (H₂) and oxygen (O₂). According to the balanced chemical equation for the formation of water, it takes 2 moles of hydrogen reacting with 1 mole of oxygen to produce 2 moles of water. The molecular weights are 2.02 g/mol for hydrogen and 32.0 g/mol for oxygen, so 2 moles of hydrogen (4.04 g) react with 1 mole of oxygen (32.0 g) to produce 2 moles (36.04 g) of water.
For the given masses, we can apply stoichiometry to find out how much water is produced. First, we convert the mass of hydrogen to moles by dividing by its molar mass:
10.54 g H₂ × (1 mol H₂ / 2.02 g H₂) = 5.22 mol H₂
Next, we do the same for oxygen:
95.10 g O₂ × (1 mol O₂ / 32.0 g O₂) = 2.97 mol O₂
Using the stoichiometry of the balanced equation (2:1 ratio of H₂:O₂), we determine the limiting reactant. In this case, hydrogen is the limiting reactant since it would require 2.97 x 2 = 5.94 moles of hydrogen to fully react with the oxygen. Therefore, all 5.22 moles of hydrogen will react.
From the stoichiometry, 2 moles of hydrogen produce 2 moles of water. Therefore, 5.22 moles of hydrogen will produce 5.22 moles of water. Finally, we convert the moles of water to grams using the molar mass of water (18.02 g/mol):
5.22 moles H₂O × 18.02 g/mol H₂O = 94.04 g H₂O
So, 94.04 grams of water will form when 10.54 g H₂ reacts with 95.10 g O₂.