M- 2/7- Ruderman

I. On the way into class today, everyone picked up Journal Page 29. It had 4 problems dealing with Limiting Reactants (L.R.) and Excess Reactants (E.R.).

Here is a sample problem:







Step-by-Step:
1. Write what you are starting with on the left, and what you are looking for on the right.
2. Figure out what kind of problem it is (this one is grams-to-grams). Also you are trying to find out what the limiting reactant is, which means you have to do this type of problem twice.
3. In a grams-to-grams problem, the first step is looking at the periodic table and finding out the mass of the known thing. In this problem the first known thing is H2. The mass of H is 1.01g. Since there is two of them, you do 1.01+1.01=2.02, and you put 2.02gH2 on the bottom so it can cancel out the 1.0gH2. On the top, you put 1 mol of H2.
4. Next, you need to look at the periodic table, because that is the next step to finding grams (see mol island). On the balanced equation, it says that there are 2 mols of H2. You would put that on bottom to cancel out the 1 molH2. On the top you would put 2 mols H2O. You would use 2 mols H20 because that is what it says in the balanced equation. You would put it on the top because gH20 is what you are looking for.
5. Then, you would have to look at the periodic table again except for the unknown thing (H2O). H is 1.01 and O is 16.00, so you would do 1.01+1.01+16.00=18.02. You would put 1 molH2O on bottom to cance lout the 2 mols of H2O on top in the previous step. Then, you would put the 18.02 on top, because that is what we are looking for.
6. Then you would type the following into your calculator:
1.0 * 2 * 18.02 [Enter]. / 2.02 [Enter]. / 2 [Enter]. and you should get 8.921.
7. You would follow the same procedure to find out the answer to the second part of this question.
8. After doing all of the work for both problems, the answers were 8.921 and 1.126. Since 1.126 is the smaller number, the O2 would be the L.R.
II.
Then Mr. Tucker gave us back Quiz 3. The answers were:
1. Balanced Equation should be: 2H2O2----> 2H2O + O2
2. 13.2gH2O
3. 3.95L O2
III.
Then Mr. Tucker further explained....
1. What L.R. are
2. How to find the L.R.
3. How to find the E.R.
4. How to find how much E.R. will be left over
5. How much product can be made
Then Mr. Tucker went over problem 1a on J.P. 29 (just as I did above in the example).
Finally, we took Quiz 4.

HOMEWORK:
1. Journal Page 29
2. Review previous notes on L.R.
3. Quiz 5 tomorow!!!