Lesson 4: Limiting Reagents Using a Balanced Equation
Limiting reagent problems that are solved using balanced equations are very much the same as what you were working with in the previous section. It's just that you now use the balanced equation to give you the relationship between the chemicals.
There is a sequence of steps that should be taken to solve these types of problems. First, write an equation for the reaction and balance it. Second, determine the mole and weight relationships among the chemicals in the reaction. Third, determine the limiting reagent in the same manner you did in the previous section. Fourth, carry out the necessary calculations using the mole and weight relationships determined from the balanced equation.
Two examples are worked through for you on the pages in this section, one with weights (ex. 4 from your workbook) and one with moles (ex. 5 from your workbook). Look through them and then try your hand at the practice problems that follow (ex. 6 from your workbook). The answers to those practice problem are at the end of this page.
Weight Example | Moles Example | Practice
Weight Example
This problem starts by giving you some information (also shown in ex. 4 in your workbook): methane (CH4) burns in air (that means it reacts with oxygen, O2) to form carbon dioxide (CO2) and also form water vapor (H2O). |
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We will need to go through a sequence of steps to solve this problem. So, the first thing we need is to write down the equation for the reaction and balance it. Everything in the question is in grams, so we don't really need to write down the mole relationship, but it helps in the next step. |
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We need the weight relationships, which are shown. A very useful thing at this
point, just to double check your math is to realize that the conservation of mass applies.
So, the total weight of reactants should be equal to the total weight of the products.
Checking that will serve as a double check on the math that you've done so far. Sixteen
plus 64 is 80 and 44 plus 36 is also 80, so that checks out. |
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Now, we need to determine the limiting reagent in the same way that you did earlier in this lesson. In this case that can be done using a rough approximation. |
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We have
20 g of methane mixed with 30 g of oxygen. What we have available is half-again as much
oxygen as we have methane. However, if you look at the weight relationship, you can see
that it takes about four times as much oxygen. So we're going to run out of oxygen. We
just don't have enough oxygen to react with all of the methane that's available. The
limiting reagent is 30 g of oxygen. |
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If you don't feel comfortable using a rough approximation, then feel free to do the calculations to determine the limiting reagent. In fact, doing the calculations as a way to check your approximating skills is useful for improving your ability to perform such calculations. |
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To answer the question, "How much excess reagent would be left
over?," we will have to know how much methane (which is the excess reagent) actually
reacts. So you need to find out how much methane reacts with 30 g of oxygen. The
difference between that and the 20 g that is available will be the amount that's left
over. The calculations are shown. |
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Next we need to find out how much carbon dioxide can be made from 30 g of oxygen and how much water can be made from 30 g of oxygen. For our purposes, let's consider the zeroes in 20 and 30 to be significant digits. The calculations are shown. |
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Moles Example
Here is a very similar kind of problem except in this case we work with moles. (It is also given in ex. 5 in your workbook.) We have different starting information but then the same questions as before. |
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Therefore, the equation to balance is the same one that we had before.
The mole relationships are shown by the coefficients: one mole of methane reacts with two
moles of oxygen to form one mole of carbon dioxide and two moles of water. |
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| Next we need to determine the limiting reagent, and carry out the calculations. The calculations are shown. Look through them to make sure you can follow along. |
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Practice
Next, take some time to work through the practice problems in ex. 6 in your workbook. With each of the sets you will need to balance the equations, determine the limiting reagent, and then answer the questions given. When you are done with that, check your answers. If you've got them all correct, great. If you have problems with any of them, take the time to sort out what it is that's causing the problem, work on that, and get it squared away.
Here are the balanced equations and the answers to the questions.
4 K + O2  2 K2O
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Yes, 0.65 mole O2 is more than enough. |
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Yes, 0.65 g O2 is more than enough. |
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O2 will be the limiting reagent and 0.59 g K2O will be formed. |
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Na2O + H2O 2 NaOH
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Na2O will be the limiting reagent and 16.0 g NaOH will be formed. |
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Na2O will be the limiting reagent because there is more than enough H2O. |
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C + 2 H2 CH4
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H2 will be the limiting reagent and 3.5 moles CH4 will be formed. |
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C will be the limiting reagent and 6.7 g CH4 will be formed. |
Make sure that you resolve any problems you might have had with these questions before moving on to the next topic.
If you want or need additional practice, there are questions at the ends of the chapters in your text that will give you some additional practice in working on the kinds of problems we have done in this lesson. You can also get some more from your instructor.