#### Limiting Reagents

Basic Assumptions and notes:
If measured amounts of two substances are reacted then we can calculate the number of moles of each reagent used
-         then the balanced equation can be used to determine which material is the limiting reagent
-         the quantity of product will depend on the limiting reagent

Sample calculation:

Copper reacts with dilute nitric acid according to the equation:

3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)

A 200.0 mL sample of 2.00 mol/L nitric acid is added to 30.48 g of copper. Determine which reactant is the limiting reagent. What volume of nitrogen monoxide, measured at 99.5 kPa and 20 oC, would be produced?

solution

determine the # of moles of acid

moles of HNO3 = 0.2000 L HNO3 x2.00 mol/L HNO3
= 0.400 mol HNO3

determine the number of moles of copper

moles of Cu = 30.48 g ÷  63.55 g/mol = 0.4796 mol Cu

determine the limiting reagent

from the equation we see that
3 mol Cu → 8 mol HNO3              using # mole Cu actually have,
0.4796 mol Cu → x mol HNO3              calculate the # of  moles of HNO3  required

x mol HNO3 = 8 mol HNO3 x 0.4796 mol Cu
3 mol Cu
= 1.279 mol HNO3

only 0.400 mol of HNO3 are available, therefore the limiting reagent is HNO3

determine the # of moles of NO

the amount of limiting reagent 0.400 mol HNO3 determines the amount of NO produced
from the balanced equation we know that the ratio is

8 mol HNO3 → 2 mol NO produced
0.400 mol HNO3 → x mol NO

x = 2 mol NO x 0.400 mol HNO3
8 mol HNO3
= 0.100 mol NO

determine the volume of NO by using the gas law

V = nRT
P

V = 0.100 mol x  8.34 kPa•L•mol–1•K–1 293 K
99.5 kPa
= 2.45 L NO