Limiting Reactants

Question

For this reaction, 14.5 g nitrogen monoxide reacts with 5.42 g oxygen gas.

nitrogen monoxide (g) + oxygen (g)  nitrogen dioxide (g)

What is the maximum mass of nitrogen dioxide that can be formed?  

What is the FORMULA for the limiting reagent? 

What mass of the excess reagent remains after the reaction is complete

Answer 1

NO + O2 → NO2

Balance:

2NO + O2 → 2NO2

14.5 g NO x (1 mol NO/30.01 g NO) x (1 mol NO2/2 mol NO) = 0.234 mol NO2

5.42 g O2 x (1 mol O2/32.00g O2) x (1 mol NO2/ 1 mol O2) = 0.169 mol NO2

Choose the reactant that produces the least amount of product: 5.42 g O2

Continue:

5.42 g O2 x (1 mol O2/32.00g O2) x (1 mol NO2/ 1 mol O2) x 46.01 g NO2/1 mol NO2) = 7.79 g NO2

Repeat for LR:

5.42 g O2 x (1 mol O2/32.00g O2) x (1 mol NO/ 1 mol O2) x (30.01 g NO/1 mol NO = 5.08 g NO used

14.5 g NO - 5.08 g NO used = 9.42 g NO left over, or 9.4 g if following rules sig fig for subtraction