#### 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