If 6.2 moles of nitrogen monoxide (no) react with 13.0 moles of oxygen gas (o2), how many moles of the product can be formed and how many moles of the excess reactant will be left over when the reaction is complete? show all of your work. unbalanced equation: no + o2 "yields" / no2

2NO + O2 → 2NO2

(a) Identify the limiting reactant

From NO:

Moles of NO2 = 6.2 mol NO * (2 mol NO2/2 mol NO) = 6.2 mol NO2

From O2:

Moles of NO2 = 13.0 mol O2 * (2 mol NO2/1 mol O2) = 26.0 mol NO2

The limiting reactant is NO, because it gives fewer moles of product.

∴ Moles of product = 6.2 mol NO2.

(b) Calculate the moles of excess reactant (O2) used

Moles of O2 = 6.2 mol NO * (1 mol O2/2 mol NO) = 3.1 mol NO2

(c) Calculate the moles of unused excess reactant (O2)

Unused = initial - used = 13.0 mol - 3.1 mol = 9.9 mol