If 16.4 g of oxygen gas react with excess hydrogen, what mass of water is produced?                                                                                        2H2(g)  +  O2(g)  ®  2H2O(g)

To solve this problem, we need to use the concept of stoichiometry, which is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here are the steps:

1. Write down the balanced chemical equation: 2H2(g) + O2(g) -> 2H2O(g)

2. Determine the molar mass of Oxygen (O2) and Water (H2O). The molar mass of O2 is approximately 32 g/mol and the molar mass of H2O is approximately 18 g/mol.

3. Convert the given mass of Oxygen to moles using its molar mass. Moles of O2 = 16.4 g / 32 g/mol = 0.5125 mol

4. Use the stoichiometric ratio from the balanced chemical equation to find the moles of water produced. From the balanced equation, we can see that 1 mol of O2 produces 2 mol of H2O. So, 0.5125 mol of O2 will produce 2 * 0.5125 mol = 1.025 mol of H2O.

5. Convert the moles of water to grams using its molar mass. Mass of H2O = 1.025 mol * 18 g/mol = 18.45 g

So, if 16.4 g of oxygen gas react with excess hydrogen, 18.45 g of water will be produced.