Last class of February!
Feb 11, 2010
at
8:17 PM
In today's class, we were put to the test as Mr. Doktor challenged us to create 2.00g of Barium Chloride as a precipitate for a grand prize of... donuts. Yes that's right, even better than good grades!! Haha. Anyways, we had to create our own procedure and were given different chemical compounds to choose from in order to create Barium Chloride for ourselves. Now that's it over and done with, the only question is: who gets the donuts?
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Mass 2 Mass!
Feb 4, 2010
at
8:06 PM
Mass to Mass EXAMPLES:
Lead (IV) nitrate reactions with 5.0 g of Potassium iodide.
Grams of Lead (IV) Nitrate are required?
Pb(NO3) + 4 KI -----> 4 KO3 + PbI45.0 g x 1 mol KI/165.9 g x 1 Pb(NO3)4/4KI x 455.2/ 1 mol Pb(NO3)4 = 3.4 gif a 100 mL solution of 2.0 M H2SO4 is neutralized by sodium hydroxide what mass of water is produced?
H2(SO4) + NaOH -----> HOH + Na2(SO4)2.0 mol/L x .1 L = .200 mol H2(SO4) x 2 HOH/1 H2(SO4) x 18.0 g/mol = 7.2 g H2OPercent Yield
The theoretical yield of the reaction is the expected *calculated* amount. The experimental amount is the actual yield% yield = actual/theoretical x 100
IE: He production of urea CO(NH2) is given by:
2NH3 + Co2 -----> CO(NH2) + H2O. 47.7 g of urea are produced when 1 mole of CO2 reacts, find the actual yield, theoretical yield and percent yield.
The actual yield is 46.7 g.
The theoretical yield: 1 mol CO2 x 1 CO(NH2)2/1 CO2 x 60.1 g/ 1 mol = 60.1 g
Percent Yield = 47.7 g/60.1 g x 100 = 79.4%
Lead (IV) nitrate reactions with 5.0 g of Potassium iodide.
Grams of Lead (IV) Nitrate are required?
Pb(NO3) + 4 KI -----> 4 KO3 + PbI45.0 g x 1 mol KI/165.9 g x 1 Pb(NO3)4/4KI x 455.2/ 1 mol Pb(NO3)4 = 3.4 gif a 100 mL solution of 2.0 M H2SO4 is neutralized by sodium hydroxide what mass of water is produced?
H2(SO4) + NaOH -----> HOH + Na2(SO4)2.0 mol/L x .1 L = .200 mol H2(SO4) x 2 HOH/1 H2(SO4) x 18.0 g/mol = 7.2 g H2OPercent Yield
The theoretical yield of the reaction is the expected *calculated* amount. The experimental amount is the actual yield% yield = actual/theoretical x 100
IE: He production of urea CO(NH2) is given by:
2NH3 + Co2 -----> CO(NH2) + H2O. 47.7 g of urea are produced when 1 mole of CO2 reacts, find the actual yield, theoretical yield and percent yield.
The actual yield is 46.7 g.
The theoretical yield: 1 mol CO2 x 1 CO(NH2)2/1 CO2 x 60.1 g/ 1 mol = 60.1 g
Percent Yield = 47.7 g/60.1 g x 100 = 79.4%
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Limiting Reactants
Feb 2, 2010
at
10:59 PM
-usually one reactant gets used up first
-one reactant is completely consumed
-this determines how much product is produced
-'Guess' which reactant is limiting, and check how much of it is required
What is thelimiting reactant when 125g of P4 reacts with 323g of Cl2 to form phosphorous tricholoride?
1. P4+6Cl2--->4PCl3
2.125g X 1mol/124g X 6mol of Cl2/1mol P4 X 71g/mol = 431g of Cl2
3. Cl2 is the limiting reactant
Deternmine the theoretical yield of the previous equation
323g X 1mol Cl2/71g X 4mol PCl3/6mol CL2 X 137.4g/1mol PCl3 = 417g
Here's a link for more info on limiting reactants:
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
-one reactant is completely consumed
-this determines how much product is produced
-'Guess' which reactant is limiting, and check how much of it is required
What is thelimiting reactant when 125g of P4 reacts with 323g of Cl2 to form phosphorous tricholoride?
1. P4+6Cl2--->4PCl3
2.125g X 1mol/124g X 6mol of Cl2/1mol P4 X 71g/mol = 431g of Cl2
3. Cl2 is the limiting reactant
Deternmine the theoretical yield of the previous equation
323g X 1mol Cl2/71g X 4mol PCl3/6mol CL2 X 137.4g/1mol PCl3 = 417g
Here's a link for more info on limiting reactants:
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
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