## Tuesday, March 8, 2011

### #stoichiometry calculations involving particles - moles - mass# by:mandy

Stoichiometry???
STOY-KEE-AHM-EH-TREE is a study of amounts of substances that are involved in reactions! lots of calculations involved. So, think Math!
what are the steps to calculationg the amounts od substances in a reaction?
1. write equation
2.balance
4.calculations!
now, lets try an example:
step 1:
Mg + HCl --> MgCl2 + H2
since this reaction is not balanced yet, we'd have to preform step 2:
Mg + 2HCl --> MgCl2 + H2
step 3:
Note that when we are making a "road map", substances CANNOT be diretly convert from grams of one compound to grams od another. Instead, we have to go through moles :)
grams (x) <--> moles (x) <--> moles (y) <--> grams (y)
final step *4:
6.0g H2 x 1 mol H2/2.02g H2 x 1 mol Mg/ 1 mol H2 x 24.3g Mg/ 1 mol Mg = 72.2g Mg
YAY! good job! that was the step by step calculation.

### 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?