1. Limiting vs. Excess Reagents (Real Stoichiometry!)

2. 2 1 1 ___ + ___  ___ 10 the PB the bread
stoichiometry = the process of calculating the exact amount of products you can make from an exact amount of reactants, or vice-versa. One of the reactants will limit the amount of each product that can be made, and the other one will be in excess.
ex: making peanut butter sandwiches: you NEED 2 pieces of bread and 1 ounce of peanut butter to make 1 sandwich.
___ ___  ___
If you are GIVEN a loaf of bread with 24 pieces and a 10 oz. jar of
peanut butter, how many sandwiches can be made? ______
Which item limited the amount of sandwiches produced? __________
Which item was in excess of the amount needed? ____________
Ex1: The production of calcium phosphate involves adding lithium phosphate to calcium nitrate in a DR reaction. If you are given g of lithium phosphate and g of calcium nitrate, how many grams of each product can be made?
Write and balance the chemical equation in the problem. 2 1 1 10
the PB
the bread

3. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
Ex1: The production of calcium phosphate involves adding lithium phosphate to calcium nitrate in a DR reaction. If you are given g of lithium phosphate and g of calcium nitrate, how many grams of each product can be made?
Write and balance the chemical equation in the problem.
Get the molar masses for each compound (you will need them later on). Put these MMs in the boxes below each compound. Ignore the coefficients in calculating molar masses.
The first major step in the L vs. E process is to determine which compound is LIMITING and which one is EXCESS. Do this by using 2 conversion tables:
GIVEN grams of reactant 1  grams of reactant 2 NEEDED.
GIVEN grams of reactant 2  grams of reactant 1 NEEDED.
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2

4. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
Ex1: The production of calcium phosphate involves adding lithium phosphate to calcium nitrate in a DR reaction. If you are given g of lithium phosphate and g of calcium nitrate, how many grams of each product can be made?
GIVEN grams of reactant 1  grams of reactant 2 NEEDED.
GIVEN grams of reactant 2  grams of reactant 1 NEEDED.
Look at the numbers you just calculated. If the needed amount is more than the given amount, then that compound is the limiting compound. If the needed amount is less than the given amount, then that compound is in excess.
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
Given
Needed
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2
Given
Needed
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
2 mol Li3PO4
g Li3PO4
= g
Li3PO4
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Li3PO4

5. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
GIVEN grams of reactant 1  grams of reactant 2 NEEDED.
GIVEN grams of reactant 2  grams of reactant 1 NEEDED.
Look at the numbers you just calculated. If the needed amount is more than the given amount, then that compound is the limiting compound. If the needed amount is less than the given amount, then that compound is in excess.
I NEED g of Ca(NO3)2, but I am only GIVEN g. Therefore, Ca(NO3)2 is my LIMITING reagent.
I NEED g of Li3PO4, but I am GIVEN g. Therefore, Li3PO4 is my EXCESS reagent.
Label the compounds “L” (for limiting) and “E” (for excess) above.
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
Given
Needed
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2 E
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2
Given
Needed
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
2 mol Li3PO4
g Li3PO4
= g
Li3PO4 L
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Li3PO4

6. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
The second major step in the L vs. E process is to determine how many grams of each product can be made from the amount of the LIMITING reagent you actually have. Do this by using another 2 conversion tables.
GIVEN grams of LIMITING reactant  grams of PRODUCT 1
GIVEN grams of LIMITING reactant  grams of PRODUCT 2
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2 E
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
2 mol Li3PO4
g Li3PO4
= g
Li3PO4 L
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Li3PO4
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
1 mol Ca3(PO4)2
g Ca3(PO4)2
= g
Ca3(PO4)2
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Ca3(PO4)2

7. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
GIVEN grams of LIMITING reactant  grams of PRODUCT 1
GIVEN grams of LIMITING reactant  grams of PRODUCT 2
7. Box your answers on step 6 because they are answering the question “How many grams of each product can be made?”
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2 E
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
2 mol Li3PO4
g Li3PO4
= g
Li3PO4 L
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Li3PO4
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
1 mol Ca3(PO4)2
g Ca3(PO4)2
= g
Ca3(PO4)2
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Ca3(PO4)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
6 mol LiNO3
68.95 g LiNO3
= g
LiNO3
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol LiNO3

8. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
The last major step in the L vs. E process is to determine whether or not the law of conservation of mass has been obeyed in your calculations. (HINT: it should have been!)
Add together the grams of products you just calculated (and boxed) in step This will be the total PRODUCT weight.
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2 E
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
2 mol Li3PO4
g Li3PO4
= g
Li3PO4 L
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Li3PO4
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
1 mol Ca3(PO4)2
g Ca3(PO4)2
= g
Ca3(PO4)2
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Ca3(PO4)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
6 mol LiNO3
68.95 g LiNO3
= g
LiNO3
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol LiNO3
59.14 g Ca3(PO4)2
g LiNO3
g TOTAL
PRODUCTS

9. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
Add together the amount of the LIMITING reagent you actually used (the # you wrote twice) and the amount of the EXCESS reagent you actually used (the needed number for the excess reagent).
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2 E
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
2 mol Li3PO4
g Li3PO4
= g
Li3PO4 L
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Li3PO4
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
1 mol Ca3(PO4)2
g Ca3(PO4)2
= g
Ca3(PO4)2
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Ca3(PO4)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
6 mol LiNO3
68.95 g LiNO3
= g
LiNO3
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol LiNO3
93.87 g Ca(NO3)2
g Li3PO4
g TOTAL
REACTANTS

10. 2 Li3PO4 + 3 Ca(NO3)2  1 Ca3(PO4)2 + 6 LiNO3
These two numbers should be the same (or very close)! That’s it! You’re done!
2 Li3PO Ca(NO3)2  1 Ca3(PO4) LiNO3
57.23 g Li3PO4
1 mol Li3PO4
3 mol Ca(NO3)2
g Ca(NO3)2
= g
Ca(NO3)2 E
g Li3PO4
2 mol Li3PO4
1 mol Ca(NO3)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
2 mol Li3PO4
g Li3PO4
= g
Li3PO4 L
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Li3PO4
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
1 mol Ca3(PO4)2
g Ca3(PO4)2
= g
Ca3(PO4)2
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol Ca3(PO4)2
93.87 g Ca(NO3)2
1 mol Ca(NO3)2
6 mol LiNO3
68.95 g LiNO3
= g
LiNO3
g Ca(NO3)2
3 mol Ca(NO3)2
1 mol LiNO3
59.14 g Ca3(PO4) g Ca(NO3)2
g LiNO g Li3PO4
g TOTAL g TOTAL
PRODUCTS REACTANTS
Mass was CONSERVED! 