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