Final answer:
Using stoichiometry, the reaction 2 H2(g) + O2(g) → 2 H2O(l) shows that oxygen gas (O2) is the limiting reactant with 3.0 moles of O2 producing a maximum of 6.0 moles of water. Multiplying the moles of water by its molar mass (18.02 g/mol) results in 108.12 grams of water that can be formed.
Step-by-step explanation:
To determine how many grams of water can be formed from 5.0 moles of H2 and 3.0 moles of O2, we must first write and balance the chemical equation for the reaction:
2 H2(g) + O2(g) → 2 H2O(l)
This equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. From the equation, the mole ratio of H2 to H2O is 1:1 and O2 to H2O is 1:2.
With 5.0 moles of H2, we could produce 5.0 moles of H2O. However, we only have 3.0 moles of O2, which would only produce 2*(3.0) = 6.0 moles of H2O. Thus, O2 is the limiting reactant and determines the maximal amount of H2O produced.
Using the molar mass of water (18.02 g/mol), we can calculate the mass of water produced:
Mass of water = moles of H2O × molar mass of H2O
Mass of water = 6.0 moles × 18.02 g/mol
Mass of water = 108.12 g
Therefore, the maximum amount of water that can be formed from the given amounts of H2 and O2 is 108.12 grams.