Unit 3 FAQ's
1. Do I need to check the chemical equation to see if it is balanced? Yes, You should always check any chemical equation for balance before beginning to weight relation calculations. A balanced equation is required by the principle of the conservation of mass which guide all calculations of the type presented in Units 3 and 4.
2. What is a "stoichiometric factor" ? A stoichiometric factor is the mole ratio derived from the stoichiometric coefficients in the balanced equation, relating the molar coefficient for the substance to be found or "unknown" divided by the molar coefficient of the given or "known" substance. In general usage the term mole ratio is used instead of the more esoteric, stoichiometric factor. For example, in the reaction:
3 NaBH4 + 4 BF3 = 3 NaBF4 + 2 B2H6 (Balanced ? Check)
The mole ratio between sodium borohydride, NaBH4 and diborane, B2H6 is 3 mol NaBH4 divided by 2 mol B2H6. Even for a reasonably simple equation, there are many possibilities for writing mole ratios relating various substances in the reaction; reactants to reactants, reactants to products and products to products.
3. If there are so many possible mole ratios for one equation, how do I know which one to use for any particular problem? Using the example reaction in Question 2, suppose you want to find how many mols of boron trifluoride are needed to react with a given number of mols of NaBH4. This problem asks how many mols of BF3 are needed,"unknown" and gives you how many mols of NaBH4 are present, "known", so you choose the mole ratio relating BF3 (unknown) to NaBH4 (known) that is 4 mols BF3/ 3 mols NaBH4. It does not matter how many substances are related by the balanced equation, a problem will focus on only two substances at a time and this will isolate the appropriate mole ratio from all the possibilities.
4. How do I find the limiting reactant ? To find the limiting reactant requires you to compare the actual mole ratio of the two reactants to the stoichiometric factor for the same reactants where the actual mole ratio and the stoichiometric factor are compared with the same substances in the numerator and denominator, respectively. If the numerator in the actual ratio is larger than the stoichiometric factor, with both ratios divided to have unit denominators, the "numerator substance" is in excess and the "denominator substance" is limiting. Conversely, if the numerator of the actual ratio is less than the numerator of the stoichiometric factor, the substance represented by the "actual numerator" is limiting.
Returning to the sodium borohydride- boron trifluoride reaction, suppose you have 100.0 grams of each reactant given and you need to find the limiting reactant. A mass to mol conversion of each specie a given converts 100 grams of each to 2.65 mols of NaBH4 and 1.48 mols of BF3. ( Check these two mass to mol calculations) Therefore the "actual mole ratio" in this case is given by
2.65 mols NaBH4 / 1.48 mols BF3 which is equivalent to 1.79 mols NaBH4 / 1.00 mol BF3
So the "unit denominator" ratio of the actual given quantities of reactants is 1.79 to 1.00 NaBH4 to BF3.
The stoichiometric factor between NaBH4 and BF3 is 3.00 mols NaBH4 / 4.00 mols BF3, setting this ratio to a unit denominator to compare with the "actual ratio", gives the result:
3 mols NaBH4 / 4 mols BF3 is equivalent to 0.75 mols NaBH4 / 1.00 mol BF3
Comparing these two ratios: 1.79 mols NaBH4 / 1.00 mols BF3 and 0.75 / 1.00, you can easily see that the actual relative mols of NaBH4 (numerator)_is larger than the required number from the stoichiometric factor, therefore the sodium borohydride is the excess reactant, you have more of it then you need to react with the BF3 present, and BF3 is the limiting reactant.
Note: Both ratio above compare NaBH4 to BF3. You must set your comparison ratios the same, in this case NaBH4 to BF3, however, you will find the same limiting reactant for the problem using the ratio BF3 to NaBH4. (Try the problem with the ratios BF3 to NaBH4 ).
5 Once I find the limiting reactant, what do I do with it? If the problem you are working only asks you to find the limiting reactant, you are done, In most cases the question continues and wants you to calculate the expected (theoretical ) yield of one of the products based on the complete reaction of the limiting reactant. In some questions, you might be asked to find how much of the excess reactant remains unreacted, as well. For some reason, many students are successful in finding the limiting reactant, but fail to use it properly in calculations involving it. In Question 4, you discovered 100 grams or 1.48 mols of BF3 is the limiting reactant, these are the quantaties you need to continue the calculation for how much B2H6 might theoretically be produced. You don't need to do anything with the "excess" reactant except to be aware that is is in excess and will not act to control or limit the amount of product produced.
Continuing this example, calculate how many grams of B2H6 could theoretically be produced form 100 grams of BF3 (You already know this is 1.48 mols ).
1.48 mols BF3 x [2 mols B2H6 / 4 mols BF3 ] = 0.74 mols B2H6 produced
Converting mols to mass: 0.74 mols B2H6 x 27.6 gm/mol = 20.42 grams of B2H6.
6. What is the difference between the theoretical yield of a reaction and the actual yield ? The theoretical yield of a reaction is the amount of some product, usually given in mass units of grams, that you would expect to get if the reaction based on a stoichiometric calculation not actually "running" the reaction in the laboratory. The actual yield is just that,it is the actual amount of product, in grams you actually produced after really running the experiment in the lab.Actual yield data comes from experimentally determined results. You can not "calculate" it.
7. How do I find the percentage yield of a reaction? To find the percentage yield of a reaction you need two pieces of data: the theoretical yield calculated from the stoichiometric equation and based on the limiting reactant and the actual yield. The percentage yield is the ration of the actual yield to the theoretical yield times 100 %. ( See text page 219 ).
Suppose you ran the reaction we have used as an example and mixed 100 grams of NaBH4 with 100 grams of BF3 and produced 16.65 grams of B2H6. What was the percentage yield of this reaction ? You know that the 100 grams (1.48 mols ) of BF3 was limiting and that the theoretical yield of B2H6 based on it was 20.42 grams. (Question 5 above) The actual yield is given as 16.65 grams after running the reaction. The percentage yield is then,
16.65 grams / 20.42 grams x 100 % = 81.5 %
Note: Dr George Wiger at Cal State, Dominguez Hills has developed a WEB site with practice problems for general chemistry including simple weight relation problems, limiting reactant cases and percentage yield.
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