1. Advanced - Stoichiometry

2. Classifying Reactions
Synthesis (combination): 2 or more reactants come together to form one product
Decomposition: 1 compound breaks down into 2 or more smaller compounds

3. Classifying Reactions
Single displacement (replacement): An uncombined element switches with another element in a compound 
Double displacement (replacement): 2 compounds “switch partners” with each other
***Combustion: involves oxygen gas as one of the reactants (it usually produces carbon dioxide and water)

5. YOU TRY!
Balance the following equations and indicate the type of reaction taking place
A) NaBr + H3PO4  Na3PO4 + HBr
B) H2CO3  H+ +CO32-
C) As + Cl2  AsCl3
Double displacement
Decomposition
Acid-base

6. Limiting Reactants
The reactant that runs out first and thus limits the amounts of products that can form is called the limiting reactant (or limiting reagent)

7. Limiting Reactant problem-solving plan
(For limiting reactant problems, the problem will give you information about two reactants)
A. Start with a balanced equation.
B. First determine which reactant will run out (which reactant is the limiting reactant)
1) You must change all starting masses into moles.
2) The trick to determine which reactant limits is to divide the moles of each reactant by the coefficient (from the balanced equation) associated with that reactant. The number that comes out the smallest indicates which reactant is the limiting one. The limiting reactant is the one that you must base all your other calculations on because it is the substance that limits how much of everything else can be made or is needed.
C. The other reactant (if there’s only two) will be the excess reactant, and some of it will be left over.
**Use molar ratios to convert the moles of the known limiting substance to the moles of other compounds (use coefficients from balanced equation) to answer more questions.

8. Limiting Reactants Fe + CuSO4  FeSO4 + Cu
If 120. g of Fe are reacted with 200. g of CuSO4, identify the limiting reagent. Which reagent is in excess?
120 g Fe x 1 mol Fe = 2.15 mol Fe / 1 = 2.15 (excess)
55.85 g Fe
200 g CuSO4 x 1 mol CuSO4 = 1.25 mol CuSO4 /1 = 1.25 (limiting)
g CuSO4

9. Limiting Reactant
If you are asked to determine the amount of excess reactant leftover:
A. Knowing which reactant limits and which is excess, use the limiting reactant to determine the moles of the excess reactant that is actually needed to do the reaction. Subtract.
B. Convert these moles into grams.

10. Limiting Reactants Fe + CuSO4  FeSO4 + Cu
If 120. g of Fe are reacted with 200. g of CuSO4, identify the limiting reagent. Which reagent is in excess? Calculate the mass of Copper formed. How much of the excess reagent is left over at the end of the reaction?
120 g Fe x 1 mol Fe = 2.15 mol Fe (excess)
55.85 g Fe
200 g CuSO4 x 1 mol CuSO4 = 1.25 mol CuSO4 (limiting)
g CuSO4
1.25 mol CuSO4 x 1 mol Cu x g Cu = 79.4 g Cu
1 mol CuSO mol Cu
Out of 2.15 mol Fe, only 1.25 mol is used so 0.9 mol Fe in excess
0.9 mol Fe g Fe = 53.0 grams in excess
1 mol Fe

11. Limiting Reactants
YOU TRY! (Ouch! There is LOTS to remember)
Which is the limiting reactant? Calculate the mass of lithium nitride formed from 56.0 g of N2 gas and 56.0 g of Li in the unbalanced reaction:
Li + N2  Li3N

12. Limiting Reactants
YOU TRY! (I think we should try another)
Methanol, CH3OH, which is used as a fuel in racing cars can be made by the reaction of carbon monoxide and hydrogen gas. Suppose 356 g of CO and 65.0 g of H2 are mixed and allowed to react. A) What mass of methanol can be produced? B) What mass of the excess reactant remains after the limiting reactant has been consumed?
CO(g) + H2(g)  CH3OH(l)

13. Yield
The maximum amount of product that can be produced is called the theoretical yield.
You can compare the calculated yield with the actual/experimental yield and find the percent yield. (how far off from the calculated value was your experiment)

14. Percent Yield Percent yield = actual yield x 100% theoretical yield
In previous you try example, 93.7 g Li3N should have formed. If this reaction actually gave 55.2 g Li3N, what is the percent yield of Li3N?
55.2 g Li3N x 100 = 58.9%
93.7 g Li3N

15. Percent Yield YOU TRY! Good Luck! TiCl4 + O2  TiO2 + 2Cl2
Suppose that 6.71 x 103 g of TiCl4 is reacted with 2.45 x 103 g of O2. Calculate the maximum mass of TiO2 that can form. If the percent yield of TiO2 is 75%, what mass is actually formed?