Stoichiometric Calculations

Importat - stoichiometric calculations make use of balanced equations to calculate quantities of reactants and products

Always balance a chemical equation!!

Example: 
Given the following balanced equation
 
            3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)
 
determine the volume of 2.50 mol/L nitric acid that would be required to react with 15.5 g of copper.
 
Solutions
 
Use the balanced equations and proceed to write mole ratio, mass ratio and volume ratio.
 
                        3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)
 
mole ratio:       3               8                         3                      2                4
 
mass ratio:     190.65 g    504.16 g        562.71 g             60.02 g       72.08 g
 
vol. ratio                                                                           44.8 L
(STP)
 
            convert 15.5 g of copper to moles
            moles of Cu = 15.5 g x 63.55 g/mol = 0.244 moles
           
            Determine the number of moles of HNO3
            from the equation we see that 3 moles Cu → 8 mole HNO3
                                                          0.244 moles Cu →x moles HNO3
 
                                                            x = 8 moles HNO3 x 0.244 moles Cu  = 0.651 moles HNO3
                                                                                3 moles Cu
 
            Determine the volume of acid, HNO3
 
            Use the formula c = n/v
 
            Find volume; v = n/c
                                   = 0.651 moles/2.50mol/L
                                    = 0.260 L or 260 mL
 
Example:
Given the balanced equation
 
            3Cu(s)  +  8HNO3(aq) →  3Cu(NO3)2(aq)  +  2NO(g)  +  4H2O(l)
 
determine the mass of copper that would be required to produce 4.00 L of nitrogen  monoxide at 102.5 kPa and 22 oC.
 
Solution:
 
Determine the number of moles of NO(g)
Rearrange the ideal gas law equation
 
            n = PV
                  RT
 
            n = _102.5 kPa x 4.00 L             = 0.167 mol NO
                 8.314 kPa L mol−1K−1295 K
 
From the balanced equation we see that
 
3 moles Cu → 2 moles NO produced
 
Determine the number of moles of Cu
 
            Moles Cu = 0.167 moles NO x 3 mol Cu = 0.251 mol Cu
                                         2 mol NO
The mass of copper is
 
            Mass of Cu =  0.251 mol Cu x 63.55 g Cu = 16.0 g Cu

 

                                             1 mol Cu