## Tuesday, March 8, 2011

### Molarity and Stoichiometry - Isabelle Cheng

MOLARITY AND STOICHIOMETRY
Do you remember molarity?!?!?!?!?!
Here are the formulas:
M = mol
L
L = mol
M
mol = M X L
Example:
You have 340 mL of 0.5 M NaBr solution. Since you know that you have a 0.5M solution, you know that you must have 0.5mol / L water. So, if 0.5 moles are in a liter, how many moles are in 0.340 liters? (340 mL / 1000)
M = mol solute / L water
0.5 = x mol solute / .340 L
x = (0.5)(.340) = 0.17
M = mol solute / L water
x M = 0.17 mol solute / (.340 + .560) L water
x M = 0.19 mol / L
Gas Stoichiometry Example:
Ammonia gas is reacted with sulfuric acid to form the important fertilizer ammonium sulfate. What mass of ammonium sulfate can be produced from 85 kL of ammonia reacting.

 _1_NH3 + _1_H2SO4 _1_(NH4)2SO4

1.
 85 kL = 3.469 mol. NH3 24.5 L/mol
2:
 3.469 mol. X _1_(NH4)2SO4 = 3.469 mol. (NH4)2SO4 _1_NH3
3.
 3.469 mol. X 114 g = 395 g mol
More exercises!

What volume of oxygen at STP is needed to completely burn 15 g of methanol (CH3OH) in a burner?

Find the molarity of a solution that contains 25.0 g of sodium hydroxide in 300.0 mL of solution.

If 20.5 mL of H3Po4 with an unknown molarity react with 25.0 mL of 0.500 M KOH, according to the above reaction, what is the molarity of H3PO4?