1. Chapter 12 Stoichiometry 12.2 Chemical Calculations
12.1 The Arithmetic of Equations
12.2 Chemical Calculations
12.3 Limiting Reagent and
Percent Yield
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2. How do manufacturers know how to make enough of their desired product?
CHEMISTRY & YOU
How do manufacturers know how to make enough of their desired product?
If chemical plants produce too much ammonia, then it might be wasted. But if too little is produced, then there might not be enough for all their customers.
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3. Writing and Using Mole Ratios
How are mole ratios used in chemical calculations?
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4. Writing and Using Mole Ratios
A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles.
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5. Types of Stoichiometry Problems to solve using the mole ratios of balanced chemical equations.
Mole-Mole Problems
Mole-Mass Problems
Mass-mass
Mass-volume
Volume-volume
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6. Mole-Mole Problems
Example:
NH4NO > N2O + 2 H2O
Find the number of moles of N2O and H2O produced from the decomposition of 2.25 mole of NH4NO3?

7. Mole-Mole Problems
The balanced chemical equation states that for every 1 mole of NH4NO3 that decomposes, 1 mole of N2O and 2 moles of water are produced.

8. Mole-Mole Problems
Use this mole ratio from the chemical equation to calculated the number of moles of N20 & H2O produced:
2.25 mol NH4NO mole N20
1mol NH4NO = 2.25 mole N20
2.25 mol NH4NO3 2 mole H20
1mol NH4NO3 = 4.5 mole H20
NH4NO > N2O + 2 H2O

9. 0.52 mol Mg 1 mol 02 2 mol Mg = .26 mol O2 Mole-Mole Problems
Magnesium burns in oxygen to produce magnesium oxide. How many moles of oxygen are needed to burn 0.52 moles of magnesium?
Write a balanced equation for this reaction:
2 Mg + O2  2 MgO
Solve:
0.52 mol Mg 1 mol 02
2 mol Mg = .26 mol O2

10. The equation for the composition of water is: 2 H2 + O2 ---> 2 H2O
Mole-Mole Problems
The equation for the composition of water is:
2 H2 + O2 ---> 2 H2O
How many moles of H2O are produced when 5.00 moles of oxygen are used?
5 moles O mole H2O
1 moles O2
= 10 moles H20

11. Writing and Using Mole Ratios
In chemical calculations, -mole ratios - are used to convert between a given number of moles of a reactant or product to moles of a different reactant or product.
These mole ratios are represented by the coefficients of a balanced chemical equation
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12. Writing and Using Mole Ratios
Mass-Mass Calculations
In the laboratory, the amount of a substance is usually determined by measuring its mass in grams.
If a given sample is measured in grams, then the mass can be converted to moles by using the molar mass.
Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown.
If it is the mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass.
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13. I. Mass-Mass Problems
Mass-Mass Calculations
In a mass-mass problem you are given the mass of one substance and ask to find the mass of another substance involved in the same reaction.

14. Mass-Mass Calculations
I. Mass-Mass Problems
Mass-Mass Calculations
All mass-mass problems can be solved by following these steps:
1. Write a balanced equation for the chemical reaction involved.
Convert the given mass of the substance given (A) to moles of substance (A) using the molar mass of the substance(A) .
Convert the moles of the substance given (A) to the moles of the substance (B) by using the mole ratio from the balanced chemical equation.
Change moles of substance (B) to the mass of substance (B) using the molar mass of the substance.

15. Summary of Mass-Mass Problem
Write a balanced equation for the chemical reaction involved.
Convert the given mass of the substance given (A) to moles of substance (A) using the molar mass of the substance(A) .
Convert the moles of the substance given (A) to the moles of the substance (B) by using the mole ratio from the balanced chemical equation.
Change moles of substance (B) to the mass of substance (B) using the molar mass of the substance.

16. Calculating the Mass of a Product
Sample Problem 12.4
Calculating the Mass of a Product
Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is:
N2(g) + 3H2(g)  2NH3(g)
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17. Change given unit to moles
Sample Problem 12.4
Calculate Solve for the unknown.
Start with the given quantity, and convert from mass to moles.
5.40 g H2 
1 mol H2
2.0 g H2
Given quantity
Change given unit to moles
Don’t forget to cancel the units at each step.
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18. Change given unit to moles
Sample Problem 12.4
Calculate Solve for the unknown.
Then convert from moles of reactant to moles of product by using the correct mole ratio. 
2 mol NH3
3 mol H2
5.40 g H2 
1 mol H2
2.0 g H2
Given quantity
Change given unit to moles
Mole ratio
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19. Change given unit to moles
Sample Problem 12.4
Calculate Solve for the unknown.
Finish by converting from moles to grams. Use the molar mass of NH3. 
2 mol NH3
3 mol H2
5.40 g H2 
1 mol H2
2.0 g H2
17.0 g NH3
1 mol NH3
Given quantity
Change given unit to moles
Mole ratio
Change moles to grams
= 31 g NH3
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20. What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
Phosphorus burns in air to produce a phosphorus oxide in the following reaction:
4P(s) + 5O2(g)  P4O10(s)
What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
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21. What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
Phosphorus burns in air to produce a phosphorus oxide in the following reaction:
4P(s) + 5O2(g)  P4O10(s)
What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
3.25 mol P4O10   = 403 g P
31.0 g P
1 mol P
1 mol P4O10
4 mol P
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22. Mass-Mass Calculations
Example: 50 g of Ammonium dichromate, (NH4)2Cr2O7, decomposes when heated strongly to generate nitrogen gas, N2, gaseous water, H2O, and the green crystalline compound chromium(III) oxide, Cr2O3.
How many grams of water are produced?

23. Write a balanced chemical equation for the reaction.
Mass-Mass Calculations
Write a balanced chemical equation for the reaction.
(NH4)2Cr2O7 N2 + 4 H2O + Cr2O3
50 g (NH4)2Cr2O7 1 mol (NH4)2Cr2O mol H2O g H20
252 g (NH4)2Cr2O mol (NH4)2Cr2O mol H20
= g H20

24. II. Mass-Volume Problems
Mass-Volume Calculations
In Mass-volume problems you are given the mass of one substance in a chemical reaction asked to find the volume of another substance- in this case a gas.

25. II. Mass-Volume Problems
Mass-Volume Calculations
Write the balanced equation.
Convert the given mass (A) into moles (A) using the molar mass of the substance
Convert the moles of the substance given (A) to the moles of the substance (B) by using the mole ratio from the balanced chemical equation.
Convert the number of moles (B) of the unknown quantity to volume (B) using the molar volume of a gas

26. Summary of Mass-Volume Calculations
Write the balanced equation.
Convert the given mass (A) into moles (A) using the molar mass of the substance
Convert the moles of the substance given (A) to the moles of the substance (B) by using the mole ratio from the balanced chemical equation.
Convert the number of moles (B) of the unknown quantity to volume (B) using the molar volume of a gas

27. II. Mass-Volume Problems
Mass-Volume Calculations
Example: 10.0 g of Zn react with HCl acid to produce zinc chloride and hydrogen gas. How many liters of hydrogen gas will be produced?
Write the balanced equation: Zn + 2 HCl  ZnCl2 + H2

28. II. Mass-Volume Problems
Mass-Volume Calculations
Convert the given mass (A) into moles (A) using the molar mass of the substance
Zn + 2 HCl  ZnCl2 + H2
10 g Zn 1 mol Zn
65.4 g Zn

29. II. Mass-Volume Problems
Mass-Volume Calculations
Convert the moles of the substance given (A) to the moles of the substance (B) by using the mole ratio from the balanced chemical equation:
Zn + 2 HCl  ZnCl2 + H2
10 g Zn 1 mol Zn mol H2
65.4 g Zn mol Zn

30. II. Mass-Volume Problems
Mass-Volume Calculations
Convert the number of moles (B) of the unknown quantity to volume (B) using the molar volume of a gas at STP (22.4 L)
Zn + 2 HCl  ZnCl2 + H2
10 g Zn 1 mol Zn mol H L H2
65.4 g Zn mol Zn 1 mol H2
= 3.4 L H2

31. Calculating the Molecules of a Product
Sample Problem 12.5
Calculating the Molecules of a Product
How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation?
2H2O(l) H2(g) + O2(g)
electricity
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32. g H2O  mol H2O  mol O2  molecules O2
Sample Problem 12.5
Analyze List the knowns and the unknown. 1
The following calculations need to be performed:
g H2O  mol H2O  mol O2  molecules O2
The appropriate mole ratio relating mol O2 to mol H2O from the balanced equation is:
1 mol O2
2 mol H2O
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33. Calculate Solve for the unknown.
Sample Problem 12.5
Calculate Solve for the unknown.
Start with the given quantity, and convert from mass to moles.
1 mol H2O
18.0 g H2O
Given quantity
Change to moles
29.2 g H2O 
Remember to also start your calculations with the given quantity, even if the given quantity is a product in the reaction.
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34. Calculate Solve for the unknown.
Sample Problem 12.5
Calculate Solve for the unknown. 2
Then, convert from moles of reactant to moles of product.
1 mol O2 
2 mol H2O
29.2 g H2O 
1 mol H2O
18.0 g H2O
Given quantity
Change to moles
Mole ratio
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35. Calculate Solve for the unknown.
Sample Problem 12.5
Calculate Solve for the unknown. 2
Finish by converting from moles to molecules.
6.02  1023 molecules O2
1 mol O2 
2 mol H2O
29.2 g H2O 
1 mol H2O
18.0 g H2O
Given quantity
Change to moles
Mole ratio
Change to molecules
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36. Calculate Solve for the unknown.
Sample Problem 12.5
Calculate Solve for the unknown. 2
Finish by converting from moles to molecules.
6.02  1023 molecules O2
1 mol O2 
2 mol H2O
29.2 g H2O 
1 mol H2O
18.0 g H2O
Given quantity
Change to moles
Mole ratio
Change to molecules
= 4.88  1023 molecules O2
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37. Evaluate Does the result make sense?
Sample Problem 12.5
Evaluate Does the result make sense? 3
The given mass of water should produce a little less than 1 mol of oxygen, or a little less than Avogadro’s number of molecules.
The answer should have three significant figures.
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38. Volume-Volume Stoichiometric Calculations
Sample Problem 12.6
Volume-Volume Stoichiometric Calculations
Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen react with an excess of nitrogen monoxide? Assume conditions are at STP.
2NO(g) + O2(g)  2NO2(g)
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39. Analyze List the knowns and the unknown.
Sample Problem 12.6
Analyze List the knowns and the unknown. 1
For gaseous reactants and products at STP, 1 mol of a gas is equal to 22.4 L.
KNOWNS
volume of oxygen = 34 L O2
2 mol NO2/1 mol O2 (mole ratio from balanced equation)
1 mol O2 = 22.4 L O2 (at STP)
1 mol NO2 = 22.4 L NO2 (at STP)
UNKNOWN
volume of nitrogen dioxide = ? L NO2
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40. Calculate Solve for the unknown.
Sample Problem 12.6
Calculate Solve for the unknown. 2
Start with the given quantity, and convert from volume to moles by using the volume ratio.
34 L O2 
1 mol O2
22.4 L O2
Given quantity
Change to moles
Did you notice that the 22.4 L/mol factors canceled out? This will always be true in a volume-volume problem.
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41. Calculate Solve for the unknown.
Sample Problem 12.6
Calculate Solve for the unknown. 2
Then, convert from moles of reactant to moles of product by using the correct mole ratio.
34 L O2  
1 mol O2
22.4 L O2
2 mol NO2
Given quantity
Change to moles
Mole ratio
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42. Calculate Solve for the unknown.
Sample Problem 12.6
Calculate Solve for the unknown. 2
Finish by converting from moles to liters. Use the volume ratio.
34 L O2   
1 mol O2
22.4 L O2
1 mol NO2
22.4 L NO2
2 mol NO2
Given quantity
Change to moles
Mole ratio
Change to liters
= 68 L NO2
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43. Evaluate Does the result make sense?
Sample Problem 12.6
Evaluate Does the result make sense? 3
Because 2 mol NO2 are produced for each 1 mol O2 that reacts, the volume of NO2 should be twice the given volume of O2.
The answer should have two significant figures.
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44. Finding the Volume of a Gas Needed for a Reaction
Sample Problem 12.7
Finding the Volume of a Gas Needed for a Reaction
Assuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO3 according to this balanced equation?
2SO2(g) + O2(g)  2SO3(g)
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45. Analyze List the knowns and the unknown.
Sample Problem 12.7
Analyze List the knowns and the unknown. 1
For a reaction involving gaseous reactants or products, the coefficients also indicate relative amounts of each gas.
You can use the volume ratios in the same way you have used mole ratios.
KNOWNS
volume of sulfur trioxide = 20.4 mL
2 mL SO3/1 mL O2 (volume ratio from balanced equation)
UNKNOWN
volume of oxygen = ? mL O2
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46. Calculate Solve for the unknown.
Sample Problem 12.7
Calculate Solve for the unknown. 2
Multiply the given volume by the appropriate volume ratio.
20.4 mL SO3  = 10.2 mL O2
1 mL O2
2 mL SO3
The volume ratio can be written using milliliters as the units instead of liters.
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47. Evaluate Does the result make sense?
Sample Problem 12.7
Evaluate Does the result make sense? 3
Because the volume ratio is 2 volumes SO3 to 1 volume O2, the volume of O2 should be half the volume of SO3.
The answer should have three significant figures.
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48. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What volume of water vapor is produced at STP by burning 501 g of methane?
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49. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What volume of water vapor is produced at STP by burning 501 g of methane?
501 g CH4 
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50. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What volume of water vapor is produced at STP by burning 501 g of methane?
501 g CH4
16.05 g CH4
1 mol CH4 
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51. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What volume of water vapor is produced at STP by burning 501 g of methane?
501 g CH4
16.05 g CH4
1 mol CH4 
2 mol H2O
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52. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What volume of water vapor is produced at STP by burning 501 g of methane?
22.4 L H2O
1 mol H2O
501 g CH4
16.05 g CH4
1 mol CH4
2 mol H2O
= L
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53. Key Concepts
In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of reactants, or between moles of products.
In a typical stoichiometric problem, the given quantity is first converted to moles. Then, the mole ratio from the balanced equation is used to calculate the moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole.
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54. Glossary Terms
mole ratio: a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles
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55. The Mole and Quantifying Matter
BIG IDEA
The Mole and Quantifying Matter
Mole ratios from the balanced equation are used to calculate the amount of a reactant or product in a chemical reaction from a given amount of one of the reactants or products.
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56. END OF 12.2
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