Question
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
How many moles of nitric acid are produced starting from 5.00 moles of NO2(g) and 2.00 moles of water?
Please help,
How many moles of nitric acid are produced starting from 5.00 moles of NO2(g) and 2.00 moles of water?
Please help,
Answers
DrBob222
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.
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.