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?