Tuesday, November 16

Today, after we were all seated, Mr. Henderson started off the day by showing us the fantastic job Tim did on his blog. He briefly reviewed what Tim covered in his blog, which is what we learned yesterday. Then we jumped right into working in our packets and working our way through stoichiometry grade school.

Today we learned about using the mole island to help us solve stoichiometry problems. We started on problem 6 of page 2 in our packets. We had to find how much H2 we would be needed to produce 22.8 g of NH3. First we found that the amount of moles in 22.8 g of NH3, which is 1.34 moles. Then we set the equation up like this-

22.8 g/mol NH3 x 1 mol NH3/17 g NH3 x 3 mol H2/2 mol NH3 x 20 g H2/ 1 mol H2 = 4.02 g H2

This shows that it requires 4.02 g of H2 to produce 22.8 G of NH3.
Mole island shows us how to get to a certain value. To get to the value you follow the arrows, and that's how we completed number 6. We went from C island, to moles C, to moles A, to A island. And the number of moles was the coefficient to the substance.

We then worked on the problems in page 3, which I have the answers to here.

1. Ti: 47.9 g/mol N2: 28.01 g/mol Ti3N4: 199.6 g/mol

2. 3.15 mol N2 x 1 mol Ti3N4/ 2 mol N2 = 1.58 mol Ti3N4

4. 2.85 mol Ti x 1 mol Ti2N4/ 3 mol Ti x 199.6 g Ti3N4/ 1 mol Ti3N4 = 189.6 g Ti3N4

And then as well on page 4.

6. 4.91 mol Ti3N4 x 2 mol N2/ 1 mol Ti3N4 x 28.01 g N2/ 1 mol N2 = 275 g N2

7. 26.3 g Ti x 1 mol Ti/ 27.9 g Ti x 1 mol Ti3N4/3 mol Ti x 199.6 g Ti3N4/1 mole Ti3N4= 36.5 g Ti3N4

8. 93.6 g N2 x 1 mol N2/ 28.01 g N2 x 1 mol Ti3N4/ 2 mol N2 x 199.6 g Ti3N4/ 1 mol Ti3N2 = 334 g Ti3N4

If you are able to do these, then you are are in the fifth grade at stoichiometry school!

To finish off the day we had a demo. The demo consisted of Mr. Henderson adding aluminum (Al) to copper chloride (CuCl2) in different amount then that which would make the entire solution balanced. The equation for Al + CuCl2 is balanced to this-

2Al(s) + 3CuCl2(aq) ---> 3Cu + 2AlCl3

Now by adding .02 moles of Al to .03 moles of CuCl2, our end result would theoretically have had all of the atoms reacting in the single replacement reaction. All we would have left is Cu and AlCl3. But Mr. Henderson added only .01 moles of Al in the first reaction. The .o1 was not enough, and we would have leftover CuCl2. In the second reaction, Mr. Henderson added .03 moles of Al to the CuCl2, and then we have leftover Al. In the first reaction, the aqueous solution left in the flask was a darker blue than the pure CuCl2, but it was not as dark as the second reaction. In the second reaction the flask had turned almost entirely brown in color. This was because of the combination of the brownish Cu and the grey AlCl3. This demo showed us that if we do not have substances in the proper amounts in proportion to the coefficients of the equation, then we will have leftover substances.

So in conclusion, today we learned about using mole island for setting up unit conversions, and we also learned about having leftover substances from a reaction.