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
3."road map"                                                               
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?