Limiting Reactant Problem

Question

if 22.2 grams of NO and 13.8 grams of O2 are used to form NO2, how many moles of excess reactant wilk be left over? 2NO (g) + O2 (g) -> 2 NO2 (g)

Answer 1

 2NO (g) + O₂ (g) ==> 2NO₂ (g) .. balanced equation

One easy way to determine the limiting reactant is to simply divide moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less will represent the limiting reactant

For NO: 22.2 g NO x 1 mol NO / 30.0 g = 0.740 moles NO (÷2->0.37)

For O₂: 13.8 g O₂ x 1 mol O₂ / 32 g = 0.431 moles O₂ (÷1->0.431)

Since 0.37 is less than 0.431, NO is the limiting reactant and O₂ is the excess reactant.

To find out how much O₂ is left over, we use the moles of NO and the stoichiometry of the reaction to find out how many moles of O₂ were used up. We then subtract that from the original amount of O₂ to find the amount of O₂ left over.

moles O₂ used: 0.740 mols NO x 1 mol O₂ / 2 mol NO = 0.370 mols O₂ used up

moles O₂ left over: 0.431 mols O₂ - 0.370 mols O₂ = 0.061 moles O₂ left over