If 21.4 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over? 2 NO (g) + O₂ (g) → 2 NO₂ (g)

Answer 1

You have posted many, many questions of the same type. I will answer this one, and hopefully you will use the information contained herein to answer the others. Or maybe some other tutor will be foolish enough to answer all of them. Not I.

Whenever you are given amounts of BOTH reactants, it is a LIMITING REACTANT problem, and you must first find which reactant is limiting. ONE way to do this (there are others) is to simply divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value comes out less represents the limiting reactant.

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

For NO: 21.4 g NO x 1 mol NO / 30.0 g = 0.713 mols NO (÷2->0.36)

For O2: 13.8 g O₂ x 1 mol O₂ / 32 g = 0.431 mols O₂ (÷1->0.43)

Since 0.36 is < 0.43, NO is the limiting reactant. We now use MOLES of NO to find moles of O₂ left over.

moles O₂ used up = 0.713 mols NO x 1 mol O₂ / 2 mol NO = 0.356 mols O₂ used

moles O₂ left over = 0.431 mols - 0.356 = 0.0750 mols O₂ left over