1. Chemical Equation CO(g) + 2H2(g)  CH3OH(l)
Describes a chemical reaction
A + B  C
A and B = reactants
C = product
CO(g) + 2H2(g)  CH3OH(l)

2. Information Given by Chemical Equations
CO(g) + 2H2(g)  CH3OH(l)
1 molecule of CO + 2 molecules of H2  one molecule of CH3OH
10 molecules of CO + 20 molecules of H2  10 molecules of CH3OH
1 mole of CO + 2 moles of H2  1 mole of CH3OH
Do you notice a relationship pattern?

3. Stoichiometry
Process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction.

4. Balancing equations
In a chemical reaction, atoms are neither created nor destroyed.
All atoms present in the reactants must be accounted for among the products.
Therefore, one needs to balance the chemical equation for the reaction.
What goes in must come out, albeit in a different form.

5. Rules for balancing equations
1. Balance out metals and polyatomic ions first
Consider polyatomic ions as a single species
2. Balance out elements that are found in more than one species last
Leave H for last: Na + H2O  NaOH + H2
3. Don’t get discouraged!
Balancing one atom can throw others off
Balance the equation afresh
4. Do NOT change the nature of the compound
To balance H2O you cannot make it H2O2
Why not?
5. Thus, put numbers in front of compounds
6. Lowest whole numbers
Incorrect: 4A + 4B  4AB
Correct: A + B  AB

6. Na(s) + H2O(l)  NaOH(aq) + H2(g)
Example
Balance:
Na(s) + H2O(l)  NaOH(aq) + H2(g)
Na is balanced, O is balanced, but H is not
 balance H  put 2 in front of NaOH
This gives us 4 H on the products side
Add 2 in front of water on reactants side
H’s are now balanced
So are O’s (2 on each side)
Na is now unbalanced
Add a 2 to Na on reactant side
Recapitulate: check equation
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
BALANCED!

7. Balance the following Al(s) + F2(g)  AlF3(s)
KClO3(s)  KCl(s) + O2(g)
KI(aq) + Pb(NO3)2(aq)  KNO3(aq) + PbI2(s)
C2H6(g) + O2(g)  CO2(g) + H2O(l)
Let’s talk about a neat trick for the above

8. Mole-Mole Relationships: Dimensional Analysis
Consider the decomposition of water:
2H2O(l)  2H2(g) + O2(g)
If we decompose 4 mol of H2O, how many moles of O2 do we get?
If we decompose 5.8 mol of H2O, how many moles of H2 do we get?

9. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
More practice
Calculate the moles of C3H8 used when 4.30 moles of CO2 are obtained.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

10. Mass Calculations 1. Balance the equation for the reaction.
Steps for Calculating the Masses of Reactants and Products in Chemical Reactions
1. Balance the equation for the reaction.
2. Convert the masses of reactants or products to moles.
3. Use the balanced equation to set up the appropriate mole ratio(s).
4. Use the mole ratio(s) to calculate the number of moles of the desired reactant or product.
5. Convert from moles to mass.

11. In other words… gramsA  molesA molesA  molesB molesB  gramsB
Do it all as a dimensional analysis problem!
See next slide

12. Example What mass of 2 should we weigh out to react with 35.0 g Al?
2Al(s) + 32(s)  2Al3(s)
gramsA  molesA:
molesA  molesB:
molesB  gramsB:
In one-step:
Avoids rounding errors too!

13. Practice
What mass of CO2 is produced when 96.1 g of C3H8 react with sufficient O2.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

14. Calculations Involving a Limiting Reactant
When there is insufficient amount of one of the reactants.
Calculate the mass required for the product in question using both reactants.
Which ever gives the least amount of product is the limiting reactant.
The mass of product obtained from using the limiting reactant is the correct amount.

15. Example
What mass of AlI3 will we obtain if we react 35.0 g Al with g I2?
2Al(s) + 3I2(s)  2AlI3(s)
35.0 g Al x (mol/ g) x (2 mol AlI3/2 mol Al) x ( g/mol) = 529 g AlI3
Versus: g I2 x (mol/ g I2) x (2 mol AlI3/3 mol I2) x ( g/mol) = 428 g AlI3
 I2 is the limiting reactant.
The mass of AlI3 obtained is 428 g.

16. Practice
Example
Calculate the mass of lithium nitride formed from 56.0 g of N2 and 56.0 g of Li in the balanced equation:
6Li(s) + N2(g)  2Li3N(s)

17. More practice
Example
Calculate the mass of aluminum sulfate formed from 50.0 g of Al and 50.0 g of H2SO4 in the balanced equation:
2Al(s) + 3H2SO4(aq)  Al2(SO4)3(s) + 3H2(g)

18. Percent Yield
Theoretical Yield: Amount of product predicted from the amounts of reactants used.
Actual Yield: Amount of product actually obtained. Actual yields are determined by doing the reaction
__Actual Yield__ X 100% = Percent Yield
Theoretical Yield

19. TiCl4(g) + O2(g)  TiO2(s) + 2Cl2(g)
Percent Yield
Consider the reaction:
TiCl4(g) + O2(g)  TiO2(s) + 2Cl2(g)
(a) Suppose 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.
(b) If 2.12 x 103 g was produced in the laboratory, what is the percent yield?