What is the maximum mass of H2O that can be produced by combining 84.4 g of each reactant?

4NH3 (g) + 5O2 (g) -----> 4NO (g) + 6H2O (g)

___________ g H2O ?

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To find the maximum mass of H2O that can be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed first and determines the maximum amount of product that can be formed.

1. Start by calculating the number of moles of each reactant using its molar mass.
Number of moles = Mass (g) / Molar mass (g/mol)

For NH3:
Number of moles of NH3 = 84.4 g / molar mass of NH3

For O2:
Number of moles of O2 = 84.4 g / molar mass of O2

2. Use the balanced chemical equation to determine the stoichiometric ratio of the reactants and product.
According to the balanced equation:
4NH3 (g) + 5O2 (g) -> 4NO (g) + 6H2O (g)

The ratio of NH3 to H2O is 4:6, and the ratio of O2 to H2O is 5:6.

3. Determine the limiting reactant.
Compare the mole ratios of NH3 to H2O and O2 to H2O. The reactant with the lower mole ratio is the limiting reactant.

4. Once you have determined the limiting reactant, use its mole ratio to calculate the moles of H2O that can be produced.

5. Finally, convert the moles of H2O to grams:
Mass of H2O = Moles of H2O * Molar mass of H2O

By following these steps, you can determine the maximum mass of H2O that can be produced by combining 84.4 g of each reactant.