1. Limiting Reactant Calculations
What happens if you don’t mix chemicals in the exact ratio of the recipe?

2. FeCl3 + 3 NaOH → Fe(OH)3 + 3 NaCl
Does adding more NaOH to a constant mass of FeCl3 always produce more product?
All FeCl3 is used up; adding more NaOH does not make any more product

3. Recipe for a ice cream fudge sundae:
2 scoops + 2 spoons + 1 cherry
How many sundae’s could be made by doubling ice cream?

4. 1 Sundae – need to double amounts of chocolate syrup and cherries also
Amount of syrup and cherries LIMITS # of Sundaes made
2 scoops of ice cream left over – out of syrup and cherries!

5. Limiting Reactant - Pictures
Mg + I2 → MgI Mg = I =
A) 1 atom of Mg + 2 molecules I2 → molecule(s) of MgI2
Limiting: Excess: 1 Mg

6. Limiting Reactant - Pictures
Mg + I2 → MgI Mg = I =
B) 3 atom of Mg + 2 molecules I2 → molecule(s) of MgI2
Limiting: Excess: 2 I2

7. X + Y → Z If stoichiometry of reaction is 1:1:
If X = Y , no limiting reactant
If X > Y, Y is limiting (determines amount of Z)
If X<Y, X is limiting (determines amount of Z)
If stoichiometry between X and Y is not 1:1, then must account for reacting ratio (can’t simple compare amounts)

8. Limiting Reactant - Pictures
2 H2 + O2 → 2 H2 O H = O =
A) 2 molecules of H2 + 2 molecules O2 → molecule(s) of H2O
Limiting: Excess: 2 H2
Link to CU Phet Reactant Leftovers

9. Recipe for a ice cream fudge sundae:
2 scoops + 2 spoons + 1 cherry
How many sundae’s could be made from

10. Amt ice cream 2 sundaes Amt chocolate syrup and cherries 1 sundae
Only 1 sundae can actually be made; Out of syrup and cherries

11. 2 N2H4 + N2O4 → 3 N2+ 4 H2O
100. g
200. g
? g
How many grams of N2 gas are formed when 100. g of N2H4 and 200. g of N2O4 are mixed? Strategy: In 2 Separate calculations, determine amounts of N2 that could be formed when each reactant is consumed. Compare amounts formed, smaller amount is actual answer (When limiting reactant runs out reaction stops).
100 g N2H4 → X g N2
200 g N2O4 → Y g N2
Small # of X vs Y is true theoretical yield

12. ) ) ( ( ( ) 2 N2H4 + N2O4 → 3 N2+ 4 H2O = 131 g N2 100. g 200. g ? g
How many grams of N2 gas are formed when 100. g of N2H4 and 200. g of N2O4 are mixed? ( ) (
1 mole N2H4 )
3 mole N2 ( )
100.0 g N2H4
28.02 g N2
32.06 g N2H4
2 mole N2H4
1 mole N2
= 131 g N2

13. ) ) ( ( ( ) 2 N2H4 + N2O4 → 3 N2+ 4 H2O = 183 g N2 100. g 200. g ? g
How many grams of N2 gas are formed when 100. g of N2H4 and 200. g of N2O4 are mixed? ( )
1 mole N2O4 ( )
3 mole N2 ( )
200.0 g N2O4
28.02 g N2
92.02 g N2O4
1 mole N2O4
1 mole N2
= 183 g N2

14. 2 N2H4 + N2O4 → 3 N2+ 4 H2O
100. g
200. g
? g
How many grams of N2 gas are formed when 100. g of N2H4 and 200. g of N2O4 are mixed?
100. g of N2H g N2
200. g of N2O g N2
Final Answer = 131 g N2 (smaller #) because N2H4 runs out before N2O4 stopping the reaction.

15. Homework 7-4, Self √ 9.6, p. 256
Li + N2 → Li3N (Unbalanced)

16. Homework 7-4, Self √ 9.6, p. 256
Li + N2 → 2 Li3N (Unbalanced)

17. Homework 7-4, Self √ 9.6, p. 256 6 Li + N2 → 2 Li3N
Calculate the mass of lithium nitride formed from 56.0 g nitrogen gas and 56.0 g of lithium.

18. ( ) ( ) ( ) ( ) ( ) ( ) Homework 7-4, Self √ 9.6, p. 256
6 Li + N2 → 2 Li3N ( ) ( ) ( )
1 mole Li
34.83 g Li3N
2 mole Li3N
56.0 g Li
6.94 g Li
6 mole Li
1 mole Li3N
93.7 g Li3N ( ) (
56.0 g N2
1 mole N2
2 mole Li3N ) ( )
34.83 g Li3N
28.02 g N2
1 mole N2
1 mole Li3N
139 g Li3N

19. Homework 7-4, Self √ 9.6, p. 256 56.0 g Li 93.7 g Li3N
56.0 g N g Li3N
Since 93.7 < 139 ; Answer = 93.7 g
Li = limiting reactant (it runs out first stopping reaction)

20. Homework 7-4, Self √ 9.7, p. 258 TiCl4 + O2 → TiO2 + 2 Cl2
6.71 x 103 g x 103 g ? g

21. Homework 7-4, Self √ 9.7, p. 258 TiCl4 + O2 → TiO2 + 2 Cl2
6.71 x 103 g x 103 g ? g
If % yield = 75%, what mass of TiO2 actually formed?

22. HW 7-4: p. 265, #65
Xe + 2 F2 → XeF4
130. g g ? g (Theoretical yield)

23. HW 7-4: p. 265, #65
Xe + 2 F2 → XeF4
130. g g ? g (Theoretical yield)
Actual yield = 145 g, calculate % yield

24. p.266, #69
Fe + S → FeS
5.25 g 12.7 g ? g