So a Limiting and Excess Reactant Walk Into a Bar. . .

. . . and they ask for a 100% yield of H20! Hahahaha! I'm so funny right?!

Ok, for real now! For the last couple of days we have been working on finding the limiting and execs reactants in a reaction, also known as stoichiometry. We were also introduced to percent yield today, and all of this will be shown now.

Ok so let's do problem 21 of chapter 12. This problem plays out as such:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. Wite the balanced chemical equation for the reaction. If a plant has 88g carbon dioxide and 64g water available for photosynthesis, determine:

1. The limiting reactant
2. The excess reactant ant the mass in excess
3. The mass of glucose produced.

Alright, so this is what the equations will look like to get the answer:

First the equation needs to be balanced, it will look like 6CO2 + 6H2O = C6H12O6 + 6O2. Now we can start with the actual calculation part.

Let's start with finding what is the limiting and execs reactants. We'll start with carbon dioxide because it comes first. The problem gives you that there is 88grams of CO2. We'll then convert the grams of carbon dioxide to mols. So, there are 44 grams for one mol of CO2 (yes it's labeled wrong in my notes). Now we can find the amount of glucose through the mol ratio between glucose and carbon dioxide. So, for every 6 CO2s there is 1 C6H12O6. The next part is to find how many grams are in one mol of C6H12O6. Putting it all together, there are 180grams of glucose per one mol. Multiply everything together, and then do proper division, and you'll get an answer of 60 grams of Glucose.

If you repeat the same process, but with water instead of carbon dioxide, you'll get 106.6 grams of glucose. This means that the carbon dioxide is the limiting reactant, and water is the execs reactant.
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Now to get the percent yield. To get the percent yield we need to use a different problem, in which we are given the actual percent yield. We'll try problem 27 out of the same chapter. It says:

If 14 grams of aluminum hydroxide is present in an  antacid tablet, determine the theoretical yield of aluminum chloride produced when the tablet reacts with stomach acid. If the actual yield of aluminum chloride from this tablet is 22 grams, what is the percent yield?
Al(OH)2 + 3HCl -> AlCl3 + 3H2O

If you do all the math and such you will get a theoretical yield of 23.9 grams of AlCl3. Our actual percent yield is 22 grams of AlCl3. To get the percent yield we need to take the actual yield divided by the theoretical yield.

That would give us: 22/23.9 = .882.
Multiply this by 100 and you get a total of 88.2% yield of Ag.   DONE!