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−1x 295 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