Lesson 3: Using Balanced Equations
Now that you know how to balance equations, you need to know what to do with them and how you can use them. The way that you will use them is to come up with the weight relationships that exist between various chemicals involved in chemical reactions.
Information from the Balanced Equation | First Example | Second Example
Third Example | Fourth Example | Fifth Example | Practice
Information from the Balanced Equation
In the following table (also shown in example 19 in your workbook), the first line gives
the word equation "magnesium plus oxygen gives magnesium oxide." The second line
gives the balanced equation "2 Mg + O2 
2 MgO." The coefficients in front of the formulas--the
"2" in front of magnesium, the "1" that's not shown in front of
oxygen, and the "2" that's shown in front of magnesium oxide--can be interpreted
in terms of how many atoms and molecules react or in terms of how many moles of
these chemicals will react with one another. Because of the way a mole is defined, if two
magnesium atoms react with one oxygen molecule, then two moles of magnesium atoms will
react with one mole of oxygen molecules. When two moles of Mg react with one mole of O2,
two moles of MgO will be formed. Once you have the balanced equation, the coefficients
immediately give you the mole relationship. Two moles of magnesium reacts with one mole of
oxygen to give two moles of magnesium oxide.
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Since you already know how to figure the formula weights of compounds, you can use that along with the number of moles to find out the weight relationships that are involved in the chemical reaction. One mole of magnesium weighs 24.3 grams; therefore, two moles of magnesium weighs 48.6 grams. Oxygen has an atomic weight of 16.0, so the formula weight for O2 is 32.0 grams. Going over to the right side of the formula weight of magnesium oxide is the sum of the atomic weight of magnesium and the atomic weight of oxygen, which is 40.3. That is the weight in grams of one mole of magnesium oxide. Thus the weight of two moles of magnesium oxide is 80.6 grams.
Note that we can determine the weight relationship in a chemical reaction just from knowing what the chemicals are. If we know the formulas of all the chemicals involved, we can write a balanced equation. From the balanced equation, we can figure out the mole relationship. From the number of moles of each chemical that are involved in the reaction and their formula weights, we can figure out the weight of each one of these.
So from the balanced equation, you can get the weight relationships. These are the same kind of weight relationship that you were using earlier in the lesson when you were working with calculations that involved the weight relationships in chemical reactions. Now, you can do that starting with a balanced equation or actually even starting with an unbalanced equation because you can balance the equation. You don't have to carry out the reaction and weigh the chemicals or be told what the weight relationship is. You can figure that out doing the very kind of thing that we just went through in this example.
The next five examples show how to set up calculations to answer a variety of questions using information from a balanced equation. All of these examples use the equation for the formation of magnesium oxide. The questions are listed in example 20 in your workbook. If you wish, try your hand at answering these questions and then check your answers by looking at the following examples. Or you can begin by following through the worked out examples to see how they are done. Remember, the mole and weight relationships in chemical reactions based on balanced equations are called stoichiometry. After you have worked through these examples, try your hand with the practice problems (also given in Exercise 21 in your workbook) and check your answers.
First Example
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We are asked, "how much oxygen," so I write down O2 with a question mark. |
= ? O2 | ||||
Next I would write down 20.0 g of magnesium because that is the starting point for the calculation. |
20.0 g Mg | = ? O2 | |||
Next we need a conversion factor that will change grams of magnesium to grams of oxygen. |
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That conversion factor has grams of magnesium on the bottom and grams of oxygen on the top. |
20.0 g Mg | x |
g O2 g Mg |
= ? O2 | |
Then I'd look back at the previous example to get the weight relationships involved here. That shows that for every 48.6 g of magnesium that reacts, 32.0 g of oxygen reacts. |
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So, 32.0 g of oxygen goes on the top and 48.6 goes down with grams of magnesium. |
20.0 g Mg | x |
32.0 g O2 48.6 g Mg |
= ? O2 | |
That sets up the calculation to find out the amount of oxygen that would be required to react with 20.0 g of magnesium. |
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It comes out to be 13.2g O2 |
20.0 g Mg | x |
32.0 g O2 48.6 g Mg |
= 13.2 g O2 | |
Second Example
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Put down magnesium oxide with a question mark. |
= ? MgO | ||||
We write down 30.0 g of magnesium as a starting point. |
30.0 g Mg | = ? MgO | |||
Then we need a conversion factor that will change from grams of magnesium (on the bottom) to grams of magnesium oxide (on the top). |
30.0 g Mg | x |
g MgO g Mg |
= ? MgO | |
Go back to the previous example to get the relationship between magnesium and magnesium oxide. |
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We get 48.6 g for magnesium (on the bottom), and 80.6 g for magnesium oxide (on the top). |
30.0 g Mg | x |
80.6 g MgO 48.6 g Mg |
= ? MgO | |
That comes out to be 49.8 g of magnesium oxide. |
30.0 g Mg | x |
80.6 g MgO 48.6 g Mg |
= 49.8 g MgO | |
Third Example
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Again we are trying to find out how much oxygen. |
= ? O2 | ||||
Our starting point for the calculation is 50.0 g of magnesium oxide. |
50.0 g MgO | = ? O2 | |||
The conversion factor we need has grams of magnesium oxide on the bottom and grams of oxygen on the top. |
50.0 g MgO | x |
g O2 g Mg |
= ? O2 | |
The numbers based on the balanced equation are 32.0 g for oxygen and 80.6 g for magnesium oxide. |
50.0 g MgO | x |
32.0 g O2 80.6 g MgO |
= ? O2 | |
Then carry out the calculation to get the answer, 19.9 g of oxygen. |
50.0 g MgO | x |
32.0 g O2 80.6 g MgO |
= 19.9 g O2 | |
Fourth Example
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We start with moles of oxygen, so the conversion factor will have moles of oxygen on the bottom. |
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| We want to find out how many moles of magnesium so that will go on the top. | 3.0 moles O2 | = ? moles Mg | |||
| Look at the relationship in the balanced equation. It says there are two moles of magnesium for every one mole of oxygen. So, two moles of magnesium on the top and one mole of oxygen on the bottom. | 3.0 moles O2 | x |
2 moles Mg 1 mole O2 |
= ? moles Mg | |
| Completing the calculation shows that we need 6.0 moles of magnesium to react with 3.0 moles of oxygen. | 3.0 moles O2 | x |
2 moles Mg 1 mole O2 |
= 6.0 moles Mg | |
I should point out a very important thing about significant digits and the mole ratios in balanced equations. When we balance equations, we do so by thinking about how many atoms of one element react with the atoms of another element. The numbers we come up with are exact relationships. When we interpret balanced equations in terms of moles, we are still dealing with an exact relationship. Consequently, when rounding off your answer, you don't have to be concerned with the number of digits in the ratio of 2 moles of magnesium over 1 mole of oxygen. The ratio is an exact ratio. The answer will be rounded off to the number of digits shown in 3.0 moles of oxygen.
Fifth Example
In this fifth example we have both grams and moles.
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We start the calculation with 3.0 moles of MgO. We need to convert that into grams of magnesium, so the conversion factor will have moles of MgO on the bottom and grams of Mg on the top. |
3.0 moles MgO | = ? g Mg | |||
From what we did in working with the balanced equation, we know that 2 moles of MgO is equivalent to 2 moles of Mg and that 2 moles of Mg weighs 48.6 g. |
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So in the conversion factor, 48.6 goes on top with grams of Mg and 2 goes on the bottom with moles of MgO. |
3.0 moles MgO | x |
48.6 g Mg 2 mole MgO |
= ? g Mg | |
Carrying out the calculation, we get 73 g of magnesium. |
3.0 moles MgO | x |
48.6 g Mg 2 mole MgO |
= 73 g Mg | |
Practice
Next I would like you to practice with several of the same kind of problems. (These are also shown in example 21 in your workbook.) There are four sets of these practice problems. Each set starts with an unbalanced equation, so you will need to balance the equation, then develop the mole and gram relationships between the chemicals, and then work out the two or three problems for each one of those equations. Take the time now to work on that. When you have finished with the questions, check your answers below. Make sure you have the answers correct, and sort out any problems you have in working out these problems. Hopefully, they will just fall right into place. If they don't, find out why and correct the problem before you go on.
Answers:
Here are the balanced equations and the answers to the questions.
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4 K + O2 2 K2O
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Na2O + H2O 2 NaOH
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C + 2 H2 CH4
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2 C2H6 + 7 O2 4 CO2 + 6 H2O
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Make sure that you resolve any problems you might have had with these questions before finishing up this lesson.