Stoichiometry; Given the balanced chemical equation: 4Al+3O2=2Al2O3

Question

Indicate the following:

1. Atomic level interpretation
2. Mole level interpretation
3. Gram level interpretation
4. In the lab, a student is given 3.5 moles of Al, how many moles of oxygen are needed to react with this amount of Al?
5. Another student is given 20. grams of Al, how many grams of oxygen are needed?
6. Given 16.5 grams of Al and excess oxygen, what is the theoretical yield (in grams)?
7. Referring to the data given in question 6, if the actual yield is 24 grams, what is the percent yield?

Answer 1

Hello, Helen,

There is a lot to answer, but I'll keep my answers short:

1. Atomic level interpretation: 4 Al atoms combine with 3 O₂ molecules to form 2 molecules of Al₂O₃.
2. Mole level interpretation: Take what I said above and use "moles" instead of atoms or molecules.
3. Gram level interpretation: To find grams, multiply the number of moles times the molar mass of the atom or molecule. Al, for example: 4 moles times 27.0g/mole = 108 grams. The mass of all the reactants should equal the summed masses of all the products (conservation of mass).
4. In the lab, a student is given 3.5 moles of Al, how many moles of oxygen are needed to react with this amount of Al?: The molar ratio is determined by the balanced equation. It takes 3 moles of oxygen to react with 4 moles of Al. So the factor is 3 moles O₂/4 moles Al. Given 3.5 moles Al, the moles of oxygen required would be (3.5 moles Al)*(3 moles O₂/4 moles Al), or 2.635 moles O₂.
5. Another student is given 20. grams of Al, how many grams of oxygen are needed?: 20 grams of Al is (20g/27g/mole Al) = 0.741 moles of Al. Use the ratio of 3/4 (from above) to arrive at the moles of oxygen (0.556 moles O₂). Convert moles O₂ into grams by multiplying times oxygen's molar mass (0.556 moles O₂)*32 g/mole = 17.8 grams.
6. Given 16.5 grams of Al and excess oxygen, what is the theoretical yield (in grams)?: Convert the Al mass into moles and take half that to find the number of moles Al₂O₃ produced. Then determine the molar mass of Al₂O₃ and multiply that by the moles that are expected to determine the grams expected. [(moles expected)*(g Al₂O₃/mole Al₂O₃)].
7. Divide the 24 grams obtained by the grams expected (from 6) to determine percent yield. If you are over 100%, something went wrong.

I hope this helps,

Bob