1. Mole Ratios Limiting Reagent Yield Gas Stoichiometry

2. Stoichiometry Chemistry is often called the science of proportions.
Stoichiometry : the study of chemical quantities based on a balanced equation.

3. Remember the mole??
Quantities in chemical equations are counted as moles.

5. Chemical Ratios
Substances in a balanced equation are always in a given proportion to each other.
1N2 + 3H2  2NH3
The proportions can be written as ratios.

6. Mole Ratios These proportions are called “mole ratios”.
1N2 + 3H2  2NH3
For Every
There are/is
Creating ratio
1N2
3H2
1mol N2/3mol H2
2NH3 N2

7. Practice Give the 6 mole ratios for _Fe + _O2 -> _Fe2O3
4Fe : 3O2 , 4Fe : 2Fe2O3
3O2 : 4Fe , 3O2: 2Fe2O3
2Fe2O3 : 4Fe 2Fe2O3 : 3O2

8. Mole Ratios
Write the 6 mole ratios found in the balanced equation below.
2N2O5  4NO2 + 1O2
2N2O5 : 4NO2
2N2O5 : 1O2
4NO2: 2N2O5
4NO2:1O2
1O2:4NO2
1O2: 2N2O5

9. The “Given”
If 2 moles of N2O5 is reacted, how many moles of NO2 will be formed?
There is always a given in a stoichiometry problem that includes
A numeric value (2)
A metric unit (moles)
A substance (N2O5)

11. Mole-Mole Calculation
If 2 moles of N2O5 is reacted, how many moles of NO2 will be formed?
1) 2 mol N2O5 x mole ratio (which one?)
2) 2 mol N2O5 x
mol N2O5 (cancel the given)
3) 2 mol N2O5 x mol NO2 (write what you want)
mol N2O5

12. Mole-Mole Calculation
Steps
Write the given
Cancel the given units/substance
Write the desired units/substance on top
Calculate: top numbers multiply, bottom numbers are divisors

13. Mole-Mole Problems
2N2O5  4NO2 + 1O2

14. Mole-Mass Problem mol A 1 mol B
How much (mass) B can be formed from a given number of moles of A?
mol A x mol B x g B = g B
mol A mol B
1) The mol-mol ratio comes from the equation
2) The mol-mass ratio comes from the periodic table

15. Mass-Mole Problem
How many moles of B can be formed from a given mass of A?
g A x 1 mol A x mol B x = mol B
g A mol A
1) The mol-mol ratio comes from the equation
2) The mol-mass ratio comes from the periodic table

16. Mass-Mass Problem
How much (mass) B can be formed from a given mass of A?
g A x 1 mol A x mol B x g B = g B
g A mol A mol B
1) The mol-mol ratio comes from the equation
2) The mol-mass ratio comes from the periodic table

17. Mole to Mole Example
How Many moles of CO2 are produced from combustion of 2.00 moles of C6H12O6?
_C6H12O6 + _O2 → _ CO2 + _ H2O
2 mol C6H12O6 6 mol CO2
1mol C6H12O6
12 moles of CO2

18. Mole to Mole Example
How Many moles of ammonia (NH3) can be produced from 8.50 mol hydrogen that reacts with nitrogen?
_H2 + _N2 → _ NH3
8.5 mol H2 2 mol NH3
3 mol H2
5.67 moles of NH3

19. Mass to Mole Example
How many moles of water can be produced by burning 325g of octane?
_C8H18 + _O2 → _ CO2 + _ H2O
325g C8H mol C8H mol H2O
114g C8H mol C8H18
25.7 moles of H2O

20. Mass to Mole Example
How many moles of CO2 can be produced by using 145 g of oxygen?
_C + _O2 → _ CO2 
145g O2 1 mol O2 1mol CO2
32g O mol O2
4.53 moles of CO2

21. Mole to Mass Example
What mass of hydrogen can be produced by reacting mol of Al with HCl?
_Al + _HCl → _ AlCl3 + _ H2
6 mol Al 3 mol H g H2
2 mol Al 1 mol H2
18g H2

22. Mass to Mass Example
If a furnace burns an amount of coal containing 125 g of FeS2, how much SO2 (an air pollutant) is produced?
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2
125g of FeS2 1 mol FeS mol SO g SO2
120g of FeS mol feS mol SO2
133 g SO2

23. Mass to Mass Example
What mass of carbon dioxide is produced by the complete combustion of 100.0g of pentane (C5H12)?
C5H12 + 8 O2 → 5 CO2 + 6H2O
100g of C5H12 1 mol C5H mol CO g CO2
72gC5H mol C5H mol CO2
305.6 g CO2

26. Limiting Reagent It can also be thought of as a limiting ingredient
Limiting reagent = a reactant that limits the amount of product that can be made
Reactant that runs out
The limiting reactant is NOT necessarily the lower amount of reactants.
It is the reactant that will produce the least amount of product
It can also be thought of as a limiting ingredient

27. Limiting Reagent Recipe: peanut butter and jelly sandwich
2Br + 3Pb + 2Je  1 Sa
There are 3 ingredients necessary to produce one sandwich
Any one of these might run
out before the other two

28. Excess Reagents Excess = plenty of the substance
Excess = the substance won’t run out
Excess = don’t include this reagent in the calculation
If you have 20 loaves of bread, 1 jar of peanut butter and 1 jar of jelly – which ingredient is in excess?

29. Limiting Reagent
To determine the limiting reagent, more than one problem must be done
Ex: You have 8 slices of bread and 9 T of peanut butter. Which is the limiting ingredient? (2Br + 3Pb + 2Je  1 Sa)
8 Br x 1Sa = 4 Sa 9 Pb x 1Sa = 3 Sa
2 Br Pb

30. Limiting Reagent
The reactant that produces the smallest amount of product is the limiting reagent.
8 Br x 1Sa = 4 Sa 9 Pb x 1Sa = 3 Sa
2 Br Pb
Limiting reagent
Smaller quantity

32. Example 1
How many grams of silver bromide AgBr, can be formed when solutions containing 100.0g MgBr2 and 100.0g of AgNO3 are mixed? What is the limiting reactant? What is the excess Reactant? What it the amount of excess reactant?
_MgBr2 + _AgNO3 -> _AgBr + _Mg(NO3)2

33. Example 2
How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0 g of chlorine? What is the limiting reagent? What is in excess? What is the mass of the excess?

34. Example 3
How many grams of barium sulfate will be formed from of barium nitrate and 150.0g of sodium sulfate? What is the limiting reactant? What is in? what is the mass of the excess reactant?
_Ba(NO3)2 +_Na2SO4->_BaSO4+_NaNO3

35. Yield
Theoretical Yield = calculated mass of product using stoichiometry
Actual Yield = mass of product found in actual LAB
Percent Yield = actual yield x 100%
theoretical yield
The higher the percent yield, the better the results

36. Percent Yield
Methanol can be produced through the reaction of CO and H2 in the presence of a catalyst.
CO(g) + 2H2(g)  CH3OH(l)
If 75.0 g of CO reacts to produce 68.4 g CH3OH, what is the percent yield of CH3OH?
79.8%

37. Percent Yield
Chlorobenzene, C6H5Cl is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants.
C6H6(l) + Cl2(g)  C6H5Cl(s) + HCl(g)
When 36.8 g of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percent yield of C6H5Cl?
73.2%

38. Percent Yield
Aluminum reacts with excess copper (II) sulfate according to the reaction below. If 1.85 g of Al react and the percent yield of Cu is 56.6%, what mass of Cu is produced?
Al(s) + CuSO4(aq)  Al2(SO4)3(aq) + Cu(s)
3.70 g