Mg(s) + O2(g) MgO(s)
Before I can answer the question, I must balance the equation. Just as numbers without units are meaningless, an unbalanced equation is insufficient. So we balance the equation:
2Mg(s) + O2(g) 2MgO(s)
Now if I have 0.145 g of Mg, how much MgO will this produce?
The key to answering this question is to understand that reactions occur on the same mole basis as the balanced stoichiometric coefficients. This equation says that 2 moles of Mg produce w moles of MgO. So how many moles of Mg did I start with?
.145 g Mg/(24.31 g/mole) = .00596 moles of Mg
So how many moles of MgO will this reaction produce?
.00596 moles Mg * (2 moles MgO/2moles Mg) = .00596 moles of MgO
which is .00596 moles of MgO *(40.31 g MgO/mole MgO) = .240 gms of MgO
Similarly we can calculate the number of grams of oxygen needed to complete the reaction since 2 moles of Mg react with 1 mole of oxygen:
.00596 moles of Mg * (1 mole of O2/2 mole Mg) = .00298 moles O2
and the grams of oxygen would be
.00298 moles O2 * (32.00 gms O2/mole O2 = .0954 g O2
Note that we had to use 32.00 for the molar mass of oxygen as a diatomic.
Now you need to go to this site to practice
Also note above that I said "needed to complete the reaction". In these calculations we are assuming that the reaction is going to completion (at least one of the reactants is completely used up). How would we know if the reaction were not completed? We could measure the amount of MgO that we get and see if it is 0.240 g in this case (the theoretical yield). However don't forget that you may not be perfect in carrying out the reaction and in recovering your product so that the reaction could go to completion and you still not measure 0.240 g of MgO because you lost some of it. But let's use this procedure to introduce a new concept: percent yield.
Let's assume that we use 0.145 g of Mg and then we measure the amount of MgO that we get to be 0.198 g (the actual yield). What is the percent yield?
Percent yield = (actual yield)/(theoretical yield) * 100
So in this case we would have a percent yield of
(0.198/0.240) * 100 = 82.5% ......... actually considered quite good for experimental error.
Let's suppose that we are making aspirin:
2 C7H6O3(s) + C4H6O3(l) 2 C9H8O4(s) + H2O(l)
where C7H6O3 is salicylic acid, C4H6O3 is acetic anhydride, and C9H8O4 is aspirin. Note that I have already balanced the equation. Now suppose that we start with 14.43 g salicylic acid and that we obtain 6.26 grams of aspirin. What is the percent yield?
First we have to calculate the theoretical yield which is the amount that we would get if the reaction went to completion (just like we did above for MgO). So first we calculate the moles of salicylic acid that we started with:
14.43g/(138 g/mol) = .105 moles of salicylic acid
Now we multiply this by the mole ratio of 2 moles of aspirin formed for 2 moles of salicylic acid to get the number of moles of aspirin:
0.105 moles salicylic acid * (2 moles of aspirin/(2 moles of salicylic acid)) = 0.105 moles aspirin
Now the theoretical yield (in grams) of aspirin would be
0.105moles aspirin * (180 g/mol) = 18.9 grams aspirin
Then the % yield that we have obtained is
6.26 g/18.9 * 100 = 33.1 %
This also is really not a bad yield for such a reaction.
Now you need to practice percent yield problems.
Now it is time to complicate this a bit! Sometimes you don't have enough of one of the reagents to react in the mole ratio indicated by the stoichiometric coefficients. If so then the reagent in short supply is called the limiting reagent.
Before we begin a chemical equation example of the limiting reagent concept, let's look at a limiting ingredient problem. Consider the following recipe written in a different form than usual:
1 cup water + 2 cups flour + 4 squares of chocolate + 2 cups of sugar + 4 oz butter + 2 eggs --> 1 cake
Now suppose that you have the following amount of each reactant: unlimited water, 12 squares of chocolate, 4 cups of sugar, 4 cups of flour, 8 oz of butter, and 3 eggs. Answer the following questions:
Now apply this reasoning that you used to answer the above questions to a "standard" chemical reaction. How do you determine if you have a reagent in short supply? You proceed exactly as you did in the cake example, except now we must work on a mole basis. You have to calculate the moles of each reagent and then compare with the mole ratios of the stoichiometric coefficients (the recipe). Let's work an example:
3 CCl4(l) + 2 SbF3(l) 3 CCl2F2(l) + 2 Sb Cl3(s)
Note that I have balanced the equation. This is a reaction for producing freon-12 (CCl2F2). Let's suppose that we start with 150 g of CCl4 and 100 g of SbF3 and see if we have a limiting reagent problem.
The correct mole ratios would be 3 moles of carbon tetrachloride to 2 moles of SbF3. So let's see what we have.
150 g CCl4/(154 g/mol) = 0.974 moles of CCl4
100 g SbF3/(179 g/mol) = 0.559 moles of SbF3
These moles would give a SbF3/CCl4 ratio of 0.559/0.974 = 0.574 whereas what we should have (from the stoichiometric ratios) is 2/3 = 0.667
So we have a problem. Which is the limiting reagent? The number 0.574 is less than 0.667 so either the numerator is too small or the denominator is too large which says that either the amount of SbF3 is either too small or the amount of CCl4 is too large. Either way of stating this says that we don't have enough SbF3 so this is the limiting reagent. This means that SbF3 will be used up before the CCl4 is used up so in calculating the amount of freon-12 formed we will have to use the SbF3 data.
So now let's calculate the amount of freon formed and the amount of CCl4 left over after the reaction is complete.
100 g SbF3/(179 g/mol) * (3 moles freon-12 produced/2 moles of SbF3) = 0.838 moles of freon-12 produced and if we multiply this by the molar mass of freon-12 which is 121 g/mol, we will get 101 grams of freon-12 formed in the reaction.
How much CCl4 will be left over after the reaction is complete?
0.559 moles of SbF3 * (3 moles CCl4/2 moles of SbF3) = 0.839 moles of CCl4 which will be used up in the reaction. We started with 0.974 moles of CCl4 so we have 0.135 moles of CCl4 left over which is
0.135 moles CCl4 * (154 g/mol) = 20.8 grams of CCl4 left over.
Here is a site to practice limiting reagent problems.
Now for the ultimate! Practice percent yield combined with limiting reagent problems.
|Now take a practice quiz to help you understand if you understand the basic concepts.|
|You must use your real name when it asks for a name.|
|The test will only submit when you have answers all of the questions correctly.|
|If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.|
Web Author: Dr. Leon L. Combs
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