Monay, November 15

Today we started class with Mr.H handing out our grades. He assured us that even though some of us may not have done well, it is still the beginning of a quarter and there will be more tests to do well on.

After that was over with, Mr.H showed us the Academic Stimulus Program. This gave students an incentive to do WebAssign assignments early to receive extra credit. Here is a chart that shows how much extra credit is rewarded for doing the WebAssign assignments.

Assignment	Stimulus?	Due Date	Conditions of Bonus Points
Stoichiometry - Set 1	No	Tues, 11/16	--
Stoichiometry - Set 2	Yes	Thurs, 11/18	Xtra 20% if 24 hrs in advance
Stoichiometry - Set 3	Yes	Mon, 11/22	Xtra 15% if 72 hrs in advance
Stoichiometry - Set 4	Yes	Tues, 11/23	Xtra 15% if 24 hrs in advance
Stoichiometry - Set 5	Yes	Tues, 11/30	Xtra 15% if 24 hrs in advance
Stoichiometry - Set 6	Yes	Wed, 12/1	Xtra 15% if 24 hrs in advance

We then moved on the the first page to learn the basics of Stoichiometry. We learned that the coefficients in a balanced chemical equation provided information about the relative amounts of reactants and products involved in the reaction. In the equation

N2 + 3 H2 → 2 NH3 we noticed that it had coefficients of 1, 3, and 2 respectively, so for this equation there has to be 1 molecule of N2 and 3 molecules of H2 to create 2 molecules of NH3. The equation must always keep a ratio of 1:3:2. Here are the answers to problem 1.

1 molecule 3 molecules 2 molecules

2 molecules 6 molecules 4 molecules

3 molecules 9 molecules 6 molecules

5 molecules 15 molecules 10 molecules

1 mole 3 moles 2 moles

2 mole 6 moles 4 moles

10 moles 30 moles 20 moles

5.4 moles 16.2 moles 10.8 moles

1.63 moles 4.90 moles 3.27 moles

We then moved on to problem two, which got a little more difficult. In this problem, we determined the relative masses by which reactants combine to form a given mass of product using molar mass. In the same equation N2 + 3 H2 → 2 NH3 N2 had a molar mass of 28g/mole, H2 had a molar mass of 2g/mol, and NH3 had a molar mass of 17g/mol. If there was 28 grams of N2 there would have to be 6 grams of H2 to create 34 grams of NH3. Notice that the sum of the masses of N2 and H2 equals NH3. Here are the answers to problem 2.

28g 6g 34g

56g 12g 68g

2800g 600g 3400g

45g 9.64g 54.54g

32.1g 6.88g 39.98g

41.53g 8.9g 50.43g