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_NH _{3} | + | _1_H _{2}SO_{4} | _1_(NH _{4})_{2}SO_{4} |

1.

2:

85 kL= 3.469 mol. NH _{3}24.5 L/mol

3.

3.469 mol. X _1_(NH_{4})_{2}SO_{4}= 3.469 mol. (NH _{4})_{2}SO_{4}_1_NH _{3}

More exercises!

3.469 mol. X114 g = 395 g mol

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?

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