What is the amount of H2O that can be produced by combining 89.7 of each reactant in the following equation:

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

A limiting reagent problem; you know that because amounts are given for BOTH reactants.

mols NH3 = grams/molar mass.
mols O2 = grams/molar mass.

Using the coefficients in the balanced equation, convert mols NH3 to mols H2O.
Do the same for mols O2.
It is likely these two values will not be the same which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Now convert mols to grams. g = mols x molar mass

To determine the amount of H2O produced by combining 89.7 grams of each reactant, we need to calculate the number of moles of each reactant and use the stoichiometry of the balanced equation to find the mole ratio between NH3 and H2O.

First, we calculate the number of moles of NH3:
moles of NH3 = mass of NH3 / molar mass of NH3
moles of NH3 = 89.7 g / (14.01 g/mol)
moles of NH3 = 6.40 mol

Next, we use the stoichiometry of the balanced equation to find the mole ratio between NH3 and H2O. From the balanced equation, we see that the mole ratio between NH3 and H2O is 6:6 or 1:1.

Since the mole ratio between NH3 and H2O is 1:1, the amount of H2O produced will be the same as the moles of NH3, which is 6.40 mol.

Finally, we convert the number of moles of H2O to grams:
mass of H2O = moles of H2O × molar mass of H2O
mass of H2O = 6.40 mol × (18.02 g/mol)
mass of H2O = 115.33 g

Therefore, by combining 89.7 grams of each reactant, 115.33 grams of H2O can be produced according to the given equation.