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# STOICHIOMETRY

Using a balanced equation as a “ tool” to solve certain problems. 2 N 2 + 3 H 2  2 NH 3. STOICHIOMETRY. 2.56. How many moles of N 2 will react with 7.68 moles of H 2 ?. Al + O 2 . 2. 4. 3. Al 2 O 3. 2. And you’ve learned how to balance them…. Download Presentation ## STOICHIOMETRY

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1. Using a balanced equation as a “tool” to solve certain problems 2 N2 + 3 H2 2 NH3 STOICHIOMETRY 2.56 How many moles of N2 will react with 7.68 moles of H2?

2. Al + O2 2 4 3 Al2O3 2 And you’ve learned how to balance them… OK, so you’ve learned how to complete equations… But what exactly is a balanced equation good for?

3. Al + O2 4 3 Al2O3 2 A balanced equation is like a recipe: it tells you how much of one substance reacts with (or is produced by) a certain amount of another substance.

4. Al + O2 4 3 Al2O3 2 This can be very helpful to a chemical company that wants to manufacture a specific compound and needs to know how much of each starting material to purchase.

5. Al + O2 4 3 Al2O3 2 BUT… the recipe is not one that is written in cups and teaspoons like a regular recipe. Nor is it written in grams: that would not work. The recipe of a balanced equation is written in moles!

6. Al + O2 4 3 Al2O3 2 The balanced equation above, for example, reads: “Four moles of aluminum reacts with three moles of oxygen to produce two moles of aluminum oxide.”

7. Al + O2 4 3 Al2O3 2 There are actually two different ways you can solve a problem like this… So… what if you were asked to determine: How many moles of O2 would react with 7.28 moles of Al?

8. Al + O2 4 3 Al2O3 2 Since 4 moles of Al reacts with 3 moles of O2 7.28 moles of Al will react with X moles of O2. 4 mol Al 7.28 mol Al = The first way is by using proportions: 3 mol O2 X mol O2 This units would all cancel out: and the equation could then be solved by cross-multiplying: 4 x X = 3 x 7.28 4 X = 21.84 X = 5.46

9. Al + O2 4 3 Al2O3 2 4 mol Al 4 mol Al 4 mol Al 4 mol Al 3 mol O2 3 mol O2 3 mol O2 3 mol O2 2 mol Al2O3 2 mol Al2O3 2 mol Al2O3 2 mol Al2O3 Every pair of substances in a balanced equation can be thought of as two potential factor label fractions: The second way to solve the problem is by using factor label. or or or

10. Al + O2 4 3 Al2O3 2 Let’s start as we always do: by writing down the given: Since we want to cancel out “mol Al,” it goes on bottom of the factor label fraction: So let’s look at the problem again: “How many moles of O2 would react with 7.28 moles of Al?” And since we’re trying to get into “mol O2,” it goes on the top of the fraction mol O2 7.28 mol Al X mol Al and the substance) (Make sure to include the units

11. Al + O2 4 3 Al2O3 2 4 3 Now find moles Al & moles O2 in the balanced equation above, and find what coefficients go with them… Those coefficients will tell you what numbers to put into the factor label fraction: So let’s look at the problem again: “How many moles of O2 would react with 7.28 moles of Al?” “4” with mol Al and “3” with mol O2. mol O2 7.28 mol Al X mol Al and the substance) (Make sure to include the units

12. Al + O2 4 3 Al2O3 2 Of course, “mol Al” cancel out… And now you calculate the answer: Start with “7.28,” then multiply by “3” because it’s on top, then divide by “4” because it’s on bottom. So let’s look at the problem again: “How many moles of O2 would react with 7.28 moles of Al?” That gives an answer of… mol O2 3 7.28 mol Al X = 5.46 mol O2 4 mol Al and the substance) (Make sure to include the units

13. For one-step problems like the one we just solved, it probably doesn’t matter, but for two-step or three-step problems, there’s no doubt about it: the factor label method will help you get the correct answer more reliably and more quickly every time! Of course, both the proportions method and the factor label method resulted in the same correct answer: 5.46 moles O2. But which method is best to use? Because of this, we are going to focus our attention on the factor label method exclusively from here on… So get used to it!

14. Al + O2 4 3 Al2O3 2 So let’s try some more problems that use the same balanced equation as we just used.

15. Al + O2 4 3 Al2O3 2 (Right now, go to the Stoichiometry Tutorial Worksheet and try to set this problem up using the factor label method. Don’t worry about whether or not it’s right, just give it a try. After you’ve given it a try, then continue.) 1. How many moles of Al are needed to produce 0.832 moles Al2O3?

16. Al + O2 4 3 Al2O3 2 Hopefully, you started off by writing down your given (including units and substance) Then you put “mol Al2O3” on bottom of the fraction. And because we’re trying to change into “mol Al,” you put “mol Al” on top of the fraction: 1. How many moles of Al are needed to produce 0.832 moles Al2O3? Then you looked at the balanced equation and saw that “4” goes with “mol Al” and “2” goes with mol Al2O3” mol Al 4 0.832 mol Al2O3 X 2 mol Al2O3

17. Al + O2 4 3 Al2O3 2 Now go ahead and fix your set-up on the worksheet if it wasn’t already correct. Then calculate the answer. Of course “mol Al2O3” cancel out. And “0.832” times “4” divided by “2” equals… Notice that the answer has been rounded to three sig figs because that’s how many sig figs there were in your given quantity “0.832.” 1. How many moles of Al are needed to produce 0.832 moles Al2O3? mol Al 4 0.832 mol Al2O3 X = 1.66 mol Al 2 mol Al2O3

18. Al + O2 4 3 Al2O3 2 (Again, go to the Stoichiometry Tutorial Worksheet and try to set-up the problem above using the factor label method. After you’ve given it a try, then continue.) 2. How many moles of Al2O3 can be produced from 15.8 moles of O2?

19. Al + O2 4 3 Al2O3 2 Hopefully, you started off by writing down your given (including units and substance) Then you put “mol O2” on bottom of the fraction. And because we’re trying to change into “mol Al2O3,” you put “mol Al2O3” on top of the fraction: 2. How many moles of Al2O3 can be produced from 15.8 moles of O2? Then you looked at the balanced equation and saw that “2” goes with “mol Al2O3” and “3” goes with “mol O2” mol Al2O3 2 15.8 mol O2 X 3 mol O2

20. Al + O2 4 3 Al2O3 2 Now go ahead and fix your set-up on the worksheet if it wasn’t already correct. Then calculate the answer. Of course “mol O2” cancel out. And “15.8” times “2” divided by “3” equals… Notice that the answer has been rounded to three sig figs because that’s how many sig figs there were in your given quantity “15.8.” 2. How many moles of Al2O3 can be produced from 15.8 moles of O2? mol Al2O3 2 15.8 mol O2 X X = 10.5 mol Al2O3 3 mol O2

21. Al + O2 4 3 Al2O3 2 (Again, go to the Stoichiometry Tutorial Worksheet and try to set-up the problem above using the factor label method. After you’ve given it a try, then continue.) 3. When 7.3 moles of O2 react, how many moles of Al react with it ?

22. Al + O2 4 3 Al2O3 2 Hopefully, you started off by writing down your given (including units and substance) Then you put “mol O2” on bottom of the fraction. And because we’re trying to change into “mol Al,” you put “mol Al” on top of the fraction: 3. When 7.3 moles of O2 react, how many moles of Al react with it ? Then you looked at the balanced equation and saw that “4” goes with “mol Al” and “3” goes with “mol O2” mol Al 4 7.3 mol O2 X 3 mol O2

23. Al + O2 4 3 Al2O3 2 Now go ahead and fix your set-up on the worksheet if it wasn’t already correct. Then calculate the answer. Of course “mol O2” cancel out. And “7.3” times “4” divided by “3” equals… Notice that the answer has been rounded to two sig figs because that’s how many sig figs there were in your given quantity “7.3.” 3. When 7.3 moles of O2 react, how many moles of Al react with it ? mol Al 4 7.3 mol O2 X = 9.7 mol Al 3 mol O2

24. Al + O2 4 3 Al2O3 2 But what if you are given grams or atoms or molecules…What would you do then? So far all the quantities that we have worked with have been in moles. Remember, the balanced equation recipe is written in moles only.

25. Al + O2 4 3 Al2O3 2 Because the recipe is written for moles, you must always work in moles. And if your given quantity is not in moles, then you must convert it into moles before you can use the balanced equation recipe!

26. Al + O2 4 3 Al2O3 2 (Again, go to the Stoichiometry Tutorial Worksheet and try to set-up the problem above using the factor label method. Hint: after you write down your given, your first step will be to change “g Al” into “mol Al,” then your second step will be like the ones we have been doing: “mol Al” to “mol Al2O3.” After you’ve given it a try, then continue.) 4. When 18.62 g of Al react how many moles of Al2O3 are produced?

27. Al + O2 4 3 Al2O3 2 Hopefully, you started off by writing down your given (including units and substance) Then you put “g Al” on bottom of the fraction. And since we need to work in moles you put “mol Al” on top. From the periodic table, we can see that 1 mole of Al weighs 27.0 g, so“1” goes with “mol Al” and “27.0” goes with “g Al.” 4. When 18.62 g of Al react how many moles of Al2O3 are produced? Now we want to cancel out “mol Al” so that goes on bottom. And we want to get into “mol Al2O3,” so that goes on top. And from the balanced equation above we see that “2” goes with “mol Al2O3” and “4” goes with “mol Al.” mol Al mol Al2O3 1 2 18.62 g Al X X 4 27.0 g Al mol Al

28. Al + O2 4 3 Al2O3 2 Now go ahead and fix your set-up on the worksheet if it wasn’t already correct. Then calculate the answer. Of course “g Al” cancel out. And so do “mol Al.” And “18.62” times “1” divided by “27.0” times “2” divided by “4” equals… 4. When 18.62 g of Al react how many moles of Al2O3 are produced? Notice that the answer has been rounded to four sig figs because that’s how many sig figs there were in your given quantity “18.62.” mol Al mol Al2O3 1 2 = 0.3448 mol Al2O3 18.62 g Al X X 4 27.0 g Al mol Al

29. Al + O2 4 3 Al2O3 2 (Again, go to the Stoichiometry Tutorial Worksheet and try to set-up the problem above using the factor label method. Hint: after you write down your given, your first step will be to change “g O2” into “mol O2,”– remember oxygen is diatomic… then your second step will be like the ones we have been doing: “mol O2” to “mol Al.” After you’ve given it a try, then continue.) 5. How many moles of Al will react with 189 g of O2?

30. Al + O2 4 3 Al2O3 2 Hopefully, you started off by writing down your given (including units and substance) Then you put “g O2” on bottom of the fraction. And since we need to work in moles you put “mol O2” on top. From the periodic table, we can see that 1 mole of O2 weighs 32.0 g, so“1” goes with “mol O2” and “32.0” goes with “g O2.” 5. How many moles of Al will react with 189 g of O2? Now we want to cancel out “mol O2” so that goes on bottom. And we want to get into “mol Al,” so that goes on top. And from the balanced equation above we see that “4” goes with “mol Al” and “3” goes with “mol O2.” mol O2 mol Al 1 4 189 g O2 X X 3 32.0 g O2 mol O2

31. Al + O2 4 3 Al2O3 2 Now go ahead and fix your set-up on the worksheet if it wasn’t already correct. Then calculate the answer. Of course “g O2” cancel out. And so do “mol O2.” And “189” times “1” divided by “32.0” times “4” divided by “3” equals… 5. How many moles of Al will react with 189 g of O2? Notice that the answer has been rounded to three sig figs because that’s how many sig figs there were in your given quantity “189.” mol O2 mol Al 1 4 189 g O2 X X = 7.88 mol Al 3 32.0 g O2 mol O2

32. Al + O2 4 3 Al2O3 2 (Again, go to the Stoichiometry Tutorial Worksheet and try to set-up the problem above using the factor label method. Hint: this time your given is already in moles, so you can go straight to the balanced equation and convert “mol O2” into “mol Al.” But don’t stop there – the question asks for atoms, so you must then convert “mol Al” into “atoms Al.”) 6. How many atoms of Al will react with 7.8 moles of O2?

33. Al + O2 4 3 Al2O3 2 Hopefully, you started off by writing down your given (including units and substance) Then you put “mol O2” on bottom of the fraction. And though we are trying to get to “atoms Al,” we must first go to “mol Al” so that goes on top. And from the balanced equation above we see that “4” goes with “mol Al” and “3” goes with “mol O2.” 6. How many atoms of Al will react with 7.8 moles of O2? Then put “mol Al” on bottom (to cancel), and “atoms Al” on top. And we know that “1” mole equals “6.02 x1023” atoms, so…. mol Al atoms Al 4 6.02x1023 7.8 mol O2 X X 1 3 mol O2 mol Al

34. Al + O2 4 3 Al2O3 2 Now go ahead and fix your set-up on the worksheet if it wasn’t already correct. Then calculate the answer. Of course “mol O2” cancel out. And so do “mol Al.” And “7.8” times “4” divided by “3” times “6.022E23” divided by “1” equals… 6. How many atoms of Al will react with 7.8 moles of O2? Notice that the answer has been rounded to two sig figs because that’s how many sig figs there were in your given quantity “7.8.” mol Al atoms Al 4 6.022x1023 = 6.3x1024 atoms Al 7.8 mol O2 X X 1 3 mol O2 mol Al

35. Al + O2 4 3 Al2O3 2 Again, go to the Stoichiometry Tutorial Worksheet and try to set-up the problem above using the factor label method. Hint: this time your given is in molecules, so you need to convert it to moles, then use the balances equation to get from moles O2 to moles Al2O3, and then you’ll have to convert those moles into grams (because that’s what the question is asking for). That’s three whole steps! 7. What mass of Al2O3 can be produced from 3.24 x 1022 molecules of O2?

36. Al + O2 4 3 Al2O3 2 Hopefully, you started off by writing down your given (including units and substance) Then you put “molecules O2” on bottom of the fraction, and moles O2 on top” 1 mol = 6.022 x 1023 molecules, so… Then “mol O2” on bottom and “mol Al2O3” on top. And from the balanced equation above we see that “2” goes with “mol Al2O3” and “3” goes with “mol O2.” 7. What mass of Al2O3 can be produced from 3.24 x 1022 molecules of O2? Then put “mol Al2O3” on bottom (to cancel), and “g Al2O3” on top. And from adding up their masses from the periodic table, we see that “1” mole of Al2O3 equals “102 g” atoms, so…. 2 102 mol O2 mol Al2O3 g Al2O3 3.24 x 1022molecules O2 1 X X X 3 1 mol O2 mol Al2O3 molecules O2 6.022 x1023

37. Al + O2 4 3 Al2O3 2 Now go ahead and fix your set-up on the worksheet if it wasn’t already correct. Then calculate the answer. Of course “molecules O2” cancel out. And so do “mol O2” and “mol Al2O3.” And “3.24E22” times “1” divided by “6.02E23” times “2” divided by “3” times “102” divided by “1” equals… 7. What mass of Al2O3 can be produced from 3.24 x 1022 molecules of O2? Notice that the answer has been rounded to three sig figs because that’s how many sig figs there were in your given quantity “3.24 x 1022.” 2 102 mol O2 mol Al2O3 g Al2O3 3.24 x 1022molecules O2 1 X X X = 3.66 g Al2O3 3 1 mol O2 mol Al2O3 molecules O2 6.02 x1023

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