In an experiment, 6.25 g of NH3 are allowed to react with 8.50 g of O2. The chemical equation for the reaction is 4NH3 + 5O2 → 4NO + 6 H2O. How many grams of water are produced? Which reactant limits the production of water? Which reactant is in excess?

Answer 1

A limiting reagent (LR) and excess reagent (ER) problem. You know that because amounts are given for BOTH reactants; besides, the problem tell you.
Convert 6.25 g NH3 to mols. mols = grams/molar mass
Do the same for O2.

Using the coefficients in the balanced equation, convert mols NH3 to mols H2O.
Do the same and convert mols O2 to mols H2O.
It is likely that thae two values for mols H2O will not agree; the correct value in LR problems is ALWAYS the smaller number AND the reagent responsible for that lower number is the LR. The other reagent is the ER.
Now convert mols H2O to grams. g = mols x molar mass.