1. Calculating Quantities in Reactions

2. Solutions to Common Problems
particles
mass
volume
Mole given
New mole using ratios
/ 6.02x1023
/ molar mass
x 6.02x1023
/ 22.4L
x molar mass
x 22.4L
Mole to mole using the ratio of coefficients

4. What You Should Expect Given : Amount of reactants
Question: how much of products can be formed.
Example
2 A + 2B C
Given 20.0 grams of A and sufficient B, how many grams of C can be produced?

5. Mole Ratios
A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

6. 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

7. 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

8. 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

9. 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.

10. 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.

11. 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.

12. Mole-Mole Problems 0.87 mol Al 6 HCl = 2.6 mol HCl
Problem: How many moles of HCl are needed to react with 0.87 moles of Al?
Step 1: Balance The Equation
Step 2: Set up the equation identifying the given and the want
0.87 mol Al HCl Al
Step 3: Calculate the new moles using the ratios of coefficients from the balanced equation
0.87 mol Al HCl = 2.6 mol HCl
2 Al

13. Mass-Mass Problems (MassgMol; MolgMol; MolgMass)
Problem: How many grams of Al can be created decomposing 9.8g of Al2O3?
Step 1: Balance The Equation
Step 2: Calculate the moles of the given (mol/g)
Step 3: Calculate the moles of the new substance using the molar ratios
Step 4: Calculate the new mass using the new moles

15. 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.

16. 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?

17. 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

18. 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

19. 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?

20. 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

21. 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

22. 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?

23. 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

24. Balance Equations Show Proportions