1. The Math of Equations Stoichiometry
the calculation of quantities in chemical reactions

2. Limiting Reactant
Calculate the amount of product that could be made for each reactant
Compare the amount that could be made to the amount of product actually made to determine which reactant is limiting
Calculate the excess reactant by calculating the amount that actually reacts and then subtracting

3. Limiting Reactant The Chemical used up first is the LIMITING REACTANT
There is always some unreacted chemical
Because you cannot measure to the last atom
So one reactant gets used up first
Once one reactant is used up, no more product can be made
The Chemical used up first is the
LIMITING REACTANT

4. Excess reactant
So……..
reactant that is not all used up in the reaction

5. 2H2 +O2 2H2O 1. Determine the limiting reactant when..
4 grams of H2 are reacted with 31g of O2.
Determine moles of products each reactant can form
The one that is limiting will form the least amount of product
4g H2 x mole H2 x 2 moles H2O = 2 moles H2O
2 g H moles H2
31g O2 x mole O2 x moles H2O = moles H2O
32 g O moles O2
Oxygen would run out first, that’s why it makes less water.
Oxygen is the limiting reactant

6. Mg + 2HCl  MgCl2 + H2
3.500 g of Mg is added to mL of M HCl
How many grams of Magnesium chloride are formed?
3.500g Mg x 1 mole x 1 mole MgCl2 x g MgCl2 = g MgCl2
mole Mg mole MgCl2
0.3000L x 1.000mole x 1 mole MgCl2 x g MgCl2 = gMgCl2
1 L mole HCl mole MgCl2
Since Mg “runs out first”, it is the limiting reactant.
What test could you run to determine there was acid left over?

7. Mg + 2HCl  MgCl2 + H2 3.5 g of Mg is added to 300.0 mL of 1.0M HCl
How many liters of STP?
Since magnesium is the limiting reactant
3.500g Mg x 1 mole x 1 mole H2 x L H2 = L H2
mole Mg mole H2

8. Mg + 2HCl  MgCl2 + H2 What is left over and how many moles?
If magnesium is the LR, then HCl is left over
3.500g Mg x 1 mole x 2 mole HCl = moles HCl used in reaction
mole Mg
0.3000L x 1.000mole HCl = moles HCl given
1 L
moles HCl (moles HCl used by Mg) = mols HCl
0.0121moles HCl = M moles HCl x 36.5g = g
0.3000L mole HCl

9. Interpreting Balanced equations
2H2O  H2 + O2
2 molecules  2 molecules molecule
2 mole  2 mole mole

10. Steps in stoichiometry
Write symbols and balance
Write the moles on top/known on bottom
Circle the known and unknown
Do Dimensional analysis problem

11. Mole – Mole problems
Ammonium nitrate decomposes to yield dinitrogen monoxide and water
How many moles of water are produced if moles of ammonium nitrate react?
Write symbols and balance
Write the moles on top/known on bottom
Circle the known and unknown
Do Dimensional analysis problem

12. practice
Bubbles of hydrogen gas and iron(III) chloride are produced when iron is dropped into hydrochloric acid
How many moles of hydrogen are produced when 14.5 moles of iron react?

13. Mole – Mass problems
Aluminum and silver nitrate react. How many moles of aluminum nitrate will be produced if 34.6 g of aluminum react?
Write symbols and balance
Write the moles on top/known on bottom
Circle the known and unknown
Do dimensional analysis problem

14. Mass-Mass
Lithium nitride reacts with water to produce ammonia and lithium hydroxide.
How many g of LiOH are produced if .38g of Li3N react?

15. Other Sodium chloride is produced when chlorine reacts with sodium
1.How many L of Chlorine are needed to make g sodium chloride?
2.How many F.U. of NaCl are produced when x 1015 atoms of Na react?

16. Law of Conservation of Matter
Prove the law of conservation of matter for
2NH3  N H2
Do Dimensional analysis problem for each reactant and product to determine the mass in g

17. % Yield (product) % Yield = Actual (experiment) x 100%
(product) Theoretical (calculation)

18. You Try 13.5g Na x 1 mol Na x 1 mol H2 x 22.4L H2 =
13.5 g of sodium metal reacts with excess hydrochloric acid. How many liters of hydrogen are produced (at STP… 22.4L / mol)
2Na + 2HCl  2 NaCl + H2
13.5g Na x 1 mol Na x 1 mol H2 x L H2 =
23.0 g Na mol Na mol H2
6.57 liters of H2 gas
Steps to solve
Predict products
Balance equation
Set up Dimensional analysis
Solve

19. 4Al + 3O2  2Al2O3 15.0g x 1 mole 27.01 g x 2 mol Al2O3 4 mole Al
15.0 g of aluminum are reacted with excess oxygen. How much oxide was produced?
4Al + 3O2  2Al2O3
15.0g x
1 mole
27.01 g
x 2 mol Al2O3
4 mole Al
x 102.g Al2O3 =
1 mol Al2O3
28.3 g aluminum oxide

20. Four Moles of H2O! 2H2 + O2 —> 2H2O Matter is Conserved
4g g = g
8g g = ??g
8g g = g
If you react 8 grams of hydrogen and 64 g of oxygen,
how much water would did you make?
Four Moles of H2O!
Or 4 x 18 g/mole = 72 grams

21. 2H2 + O2 —> 2H2O 2 moles of H2 = how many grams?
How many grams of water are produced from 2 moles of hydrogen gas and 1 mole of oxygen gas?
2H O2 —> H2O
4g g g
2 moles of H2 = how many grams?
2 mole x 2 atoms x 1 g/mole = 4g
1 mole of O2 = how many grams?
2 atom x g/mole = 32 g
2 mole of H2O = how many grams
2 mole x 18 g/mole = 36g