In short, when 1.00 g of hydrogen and 1.00 g of oxygen react, 0.5625 g of water is formed and 0.875 g of hydrogen remains unreacted, as oxygen is the limiting reactant.
To find the amount of water produced when hydrogen (H₂) and oxygen (O₂) combine, we must consider the balanced chemical equation:
2 H₂(g) + O₂(g) → 2 H₂O(l). This indicates that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water. Since hydrogen and oxygen are both diatomic molecules in their natural gaseous states, we must also consider their molar masses.
Hydrogen has a molar mass of approximately 2 grams per mole (1 gram of H₂ is 0.5 mole), and oxygen has a molar mass of approximately 32 grams per mole (1 gram of O₂ is 1/32 mole). When we have limited quantities, the reactant present in the lowest molar amount is the limiting reactant, which in this case is oxygen. Applying stoichiometry from the balanced equation, it shows that 0.5 moles of hydrogen would react completely with 0.25 moles of oxygen to form water, but we have only 1/32 moles of oxygen available, which requires 1/16 moles of hydrogen. This means some hydrogen will remain unreacted.
Using the stoichiometry from the balanced equation, 1/32 moles of oxygen will react with 1/16 moles of hydrogen to produce 1/32 moles of water. Since the molar mass of water (H₂O) is 18 grams per mole, the mass of water produced will be 1/32 moles times 18 grams/mole, which equals 0.5625 grams.
Therefore, in theory, 0.5625 grams of water is formed, and since 1/16 moles of hydrogen corresponds to 1/16 moles * 2 grams/mole = 0.125 grams, there will be 1.00 - 0.125 = 0.875 grams of hydrogen remaining.