1. STOICHIOMETRY TUTORIAL
Paul Gilletti 1

2. Instructions: This is a work along tutorial
Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps.
Get a pencil and paper, a periodic table and a calculator, and let’s get to work. 2

3. (1-2-3) General Approach For Problem Solving:
1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?)
2. Determine what is given and the UNITS.
3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired. 3

4. Table of Contents: Click on each tab to view problem types.
View Complete Slide Show
Sample problem 1
Sample problem 2
Converting grams to moles
Mole to Mole Conversions
Gram-Mole and Gram-Gram Problems
Solution Stoichiometry Problems
Limiting/Excess/ Reactant and Theoretical Yield Problems : 4

5. Sample problem for general problem solving.
Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race.
5280 ft = 1 mile; 12 inches = 1 ft
1. What is the goal and what units are needed?
Goal = ______ inches
2. What is given and its units?
10 miles
Units match
3. Convert using factors (ratios).
10 miles = inches
633600
Goal
Given
Convert
Menu

6. This is the same as putting k over k
Sample problem #2 on problem solving.
A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine its speed in kilometers per second using the following conversion factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; k = 1 x 103; c = 1 x 10-2; 1 hr =60 min; 1 min = 60 s
Given
Goal
= km s
0.020
This is the same as putting k over k
c cancels c
m remains
Units Match!

7. Use the molar mass to convert grams to moles.
Converting grams to moles.
Determine how many moles there are in 5.17 grams of Fe(C5H5)2.
Given
Goal
units match
5.17 g Fe(C5H5)2
= moles Fe(C5H5)2
0.0278
Use the molar mass to convert grams to moles.
Fe(C5H5)2
2 x 5 x =
2 x 5 x =
1 x =

8. coefficients give MOLAR RATIOS
Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + Ba(OH)2 → 2H2O BaCl2 1 1
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

9. Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g) → 4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g) → 4NO2(g) + O2(g)
4.3 mol
? mol
Units match
4.3 mol N2O5
= moles NO2
8.6
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g) → 4NO2(g) + O2(g)
4.3 mol
? mol
4.3 mol N2O5
= mole O2
2.2

10. gram ↔ mole and gram ↔ gram conversions
When N2O5 is heated, it decomposes:
2N2O5(g) → 4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g) → 4NO2(g) + O2(g)
? moles
210g
Units match
210 g NO2
= moles N2O5
2.28
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g) → 4NO2(g) + O2(g)
? grams
75.0 g
75.0 g O2
= grams N2O5
506

11. First write a balanced equation.
Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?
Al(s) HCl(aq)  AlCl3(aq) H2(g)
First write a balanced equation.

12. Now let’s get organized. Write the information below the substances.
Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?
Al(s) HCl(aq)  AlCl3(aq) H2(g)
3.45 g
? grams
Now let’s get organized. Write the information below the substances.

13. gram to gram conversions
Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?
Al(s) HCl(aq)  AlCl3(aq) H2(g)
3.45 g
? grams
Units match
3.45 g Al
= g AlCl3
17.0
Let’s work the problem.
We must always convert to moles.
Now use the molar ratio.
Now use the molar mass to convert to grams.

15. When working problems, it is a good idea to change M into its units.
Molarity
Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M)
When working problems, it is a good idea to change M into its units.

17. What type of problem(s) is this?
Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form mL solution. A 10.0 mL portion of the solution is then used to prepare mL of solution. Determine the molarity of the final solution.
What type of problem(s) is this?
Molarity followed by dilution.

18. Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form mL solution. A 10.0 mL portion of the solution is then used to prepare mL of solution. Determine the molarity of the final solution.
1st:
3.73 g
= mol L
0.140
200.0 x 10-3 L
molar mass of AlCl3
dilution formula
M1V1 = M2V2
2nd:
(0.140 M)(10.0 mL) = (? M)(100.0 mL)
final concentration
M = M2

20. Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?
H2SO4(aq) NaHCO3 → 2H2O(l) + Na2SO4(aq) + 2CO2(g)

21. Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?
H2SO4(aq) NaHCO3 → 2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 mL
? g
Our Goal
6.0 M =
Look!
A conversion factor!

22. Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?
H2SO4(aq) NaHCO3 → 2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 mL
? g
Our Goal
6.0 M =
NaHCO3
2 mol
H2SO4
50.0 mL
NaHCO3
84.0 g
= g NaHCO3
50.4
1 mol
H2SO4
mol
NaHCO3

24. Solution Stoichiometry:
Determine how many mL of M NaOH solution are needed to neutralize 35.0 mL of M H2SO4 solution.
____NaOH + ____H2SO4  ____H2O ____Na2SO4
First write a balanced
Equation.

25. this to save on the number of conversions
Solution Stoichiometry:
Determine how many mL of M NaOH solution is needed to neutralize 35.0 mL of M H2SO4 solution.
____NaOH + ____H2SO4  ____H2O ____Na2SO4
0.102 M
35.0 mL
? mL
Our Goal
Since 1 L = 1000 mL, we can use
this to save on the number of conversions
Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound.

26. Now let’s get to work converting.
Solution Stoichiometry:
Determine how many mL of M NaOH solution is needed to neutralize 35.0 mL of M H2SO4 solution.
____NaOH + ____H2SO4  ____H2O ____Na2SO4
0.102 M
35.0 mL
? mL
Units Match
shortcut
H2SO4
35.0 mL
H2SO4
0.125 mol
1000 mL
NaOH
2 mol
1 mol
H2SO4
1000 mL NaOH
0.102 mol NaOH
= mL NaOH
85.8
Now let’s get to work converting.

28. Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
1st write out
a balanced chemical
equation

29. Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
2HCl(aq) Ba(OH)2(aq) → 2H2O(l) + BaCl2
0.40 M
47.1 mL
0.75 M
? mL
Units match
HCl
2 mol
Ba(OH)2
47.1 mL
HCl
1000 mL
176
= mL HCl
0.40 mol
HCl
1 mol
Ba(OH)2

32. First write a balanced chemical reaction.
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution. If it took mL of M hydrochloric acid to neutralize mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?
____HCl(aq) ____Ba(OH)2(aq)  ____H2O(l) ____BaCl2(aq)
23.28 mL
25.00 mL
0.135 mol L
? mol L
First write a balanced chemical reaction.

33. Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution. If it took mL of M hydrochloric acid to neutralize mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?
____HCl(aq) ____Ba(OH)2(aq)  ____H2O(l) ____BaCl2(aq)
23.28 mL
25.00 mL
Units match on top!
0.135 mol L
? mol L
= mol Ba(OH)2
L Ba(OH)2
0.0629
25.00 x L
Ba(OH)2
Units Already Match on Bottom!

35. We must first write a balanced equation.
Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of M HNO3. Determine the molarity of the Ca(OH)2 solution.
We must first write a balanced equation.

36. Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of M HNO3. Determine the molarity of the Ca(OH)2 solution.
Ca(OH)2(aq) HNO3(aq) → 2 2
H2O(l)
+ Ca(NO3)2(aq)
48.0 mL
19.2 mL
? M
0.385 M
HNO3
19.2 mL =
mol(Ca(OH)2)
L (Ca(OH)2)
0.0770
48.0 x 10-3L
units match!

38. 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
First copy down the the BALANCED equation!
Now place numerical the information below the compounds.

39. Two starting amounts? Where do we start?
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15 mol
0.10 mol
? moles
Hide
one
Two starting amounts? Where do we start?

40. 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15 mol
0.10 mol
Hide
? moles
Based on:
KO2
0.15 mol KO2
= mol O2
0.1125

41. 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15 mol
Hide
0.10 mol
? moles
Based on:
KO2
0.15 mol KO2
= mol O2
0.1125
Based on: H2O
0.10 mol H2O
= mol O2
0.150

42. 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
0.15 mol
0.10 mol
? moles
Based on:
KO2
0.15 mol KO2
= mol O2
0.1125
It was limited by the
amount of KO2.
Based on: H2O
0.10 mol H2O
= mol O2
0.150
H2O = excess (XS) reactant!
What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant?

43. Theoretical yield vs. Actual yield
Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.
Theoretical yield = 19.5 g based on limiting reactant
Actual yield = 12.3 g experimentally recovered

44. Limiting/Excess Reactant Problem with % Yield
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
If a reaction vessel contains g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
? g
120.0 g
Hide one
47.0 g
Based on:
KO2
120.0 g KO2
= g O2
40.51

45. Limiting/Excess Reactant Problem with % Yield
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
If a reaction vessel contains g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
? g
120.0 g
47.0 g
Hide
Based on:
KO2
120.0 g KO2
= g O2
40.51
47.0 g H2O
Based on:
H2O
= g O2
125.3
Question if only 35.2 g of O2 were recovered, what was the percent yield?

46. 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g
If a reaction vessel contains g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?
4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)
? g
120.0 g
47.0 g
Based on:
KO2
120.0 g KO2
= g O2
40.51
47.0 g H2O
Based on:
H2O
= g O2
125.3
Determine how many grams of Water were left over.
The Difference between the above amounts is directly RELATED to the XS H2O.
= g of O2 that could have been formed from the XS water.
84.79 g O2
= g XS H2O
31.83

48. After you have worked the problem, click here to see
Try this problem (then check your answer):
Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution.
After you have worked the problem, click here to see
setup answer