3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

Question

How many moles of nitric acid are produced starting from 5.00 moles of NO2(g) and 2.00 moles of water?

Please help,

Answer 1

This is a limiting reagent (LR) problem since an amount is given for BOTH reactants.
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

Starting with 5.00 mols NO2 and all of the water you need will produce 5.00 x (2 mol HNO3/3 mols NO2) = 5*2/3 = about 3.3

Starting with 2.00 mols H2O and all of the NO2 needed will produce 2.00 x (2 mols HNO3/1 mol H2O) = 2.00 x 2/1 = 4

You see the number of mols is different which obviously can't be correct. The correct value in LR problems is ALWAYS the smaller value.