Identifying the Limiting Reagent:
To determine the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation. The balanced chemical equation for the reaction is:

2H2 + O2 → 2H2O

First, we need to calculate the number of moles of each reactant.

Calculating the moles of H2:
Given mass of H2 = 10g
Atomic mass of H2 = 2g/mol

Number of moles of H2 = mass / atomic mass = 10g / 2g/mol = 5 mol

Calculating the moles of O2:
Given mass of O2 = 18g
Atomic mass of O2 = 16g/mol

Number of moles of O2 = mass / atomic mass = 18g / 16g/mol = 1.125 mol

Comparing the moles:
From the balanced chemical equation, we can see that 2 moles of H2 reacts with 1 mole of O2 to produce 2 moles of H2O.

Ratio of moles of H2 to O2 = 5 mol / 1.125 mol ≈ 4.44

Since the ratio is greater than 2, it means that there is an excess of H2 and O2 will be the limiting reagent in this reaction.

Calculating the amount of H2O produced:
Now that we know O2 is the limiting reagent, we can use its number of moles to calculate the amount of H2O produced.

From the balanced chemical equation, we know that 1 mole of O2 reacts to produce 2 moles of H2O.

Number of moles of H2O = 2 * 1.125 mol = 2.25 mol

Converting moles to grams:
To calculate the amount of H2O produced in grams, we need to multiply the number of moles by the molar mass of H2O.

Molar mass of H2O = 2 * atomic mass of H + atomic mass of O = 2 * 1g/mol + 16g/mol = 18g/mol

Mass of H2O produced = number of moles * molar mass = 2.25 mol * 18g/mol = 40.5g

Therefore, 40.5 grams of H2O will be produced when 10 grams of H2 reacts with 18 grams of O2.