C₈H₁₀(l) + O₂(g) + NH₃(g) → C₈H₄N₂(s) + H₂O(l). How many grams of water would be produced by the complete ammoxidation of 22.0 moles of o-xylene?

Answer 1

First we need a correctly balanced equation:

C₈H₁₀(l) + 4 O₂(g) + 2NH₃(g) → C₈H₄N₂(s) + 8H₂O(l) ... balanced equation

22.0 mol o-xylene x 8 mols H₂O / 1 mol o-xylene x 18 g H₂O/mol = 3170 g H₂O (3 sig.figs.)