Tuesday, May 17, 2011

Exam Review #21

a) (8.418x10^24) atoms X 1 mol K/ 6.022x10^23 atoms = 13.98 mol K

4(13.98) = 55.92 mol K2S

55.92 mol K2S X 110.09g/1 mol K2S = 6156 g K2S

b)38.0L CCL4 X 1 mol CCL4 / 22.4 L CCL4 =1.70 mol CCL4

c) 20.8L CO X 1 mol CO / 22.4 L C) = .929 mol CO

.929 mol CO / 2 = .465 mol 02 = 2.79 x 10^23 atoms O2


  1. on a I got 770.7 g, I dont know why you multiplied the 13.98 mol K by 4. I think the factor would be mol K2S/2mol K.

    On b, the problem asks for chloride atoms, not mol CCl4. However your calculation can help you solve for the desired answer as well. You just need to convert the figure you have into Cl atoms by multiplying by 4 mol Cl/mol CCl4 and then use avagadro's number to find the number of Cl ions in the moles of Cl you calcualte.

    On c, the problem asks for grams of Oxygen gas, and I am confused as to why you decided to multiply by 2 right off the bat instead of using the conversion factor for moles of gas at STP (22.4 L/mol)

  2. a) I got 770.7 grams of K2S. Your first step was correct. After you convert the atoms of K to moles of K, convert the moles of K into moles of K2S and then lastly convert into grams of K2S using the molar mass of 110.27 g/mol of K2S in order to receive the grams of K2S.

    b) You were on the right track again, but you forgot to include the last two steps which were to multiply by 4 mol Cl/ 1 mole CCl4 and by 6.022 X 10^23 atoms Cl/ 1 mole Cl. Your answer should be 4.09 X 10^24 atoms Cl.

    c)On this problem you might want to try it a different way. I would start with 20.8 L CO and multiply by 1 mole CO/22.4 L CO, 1 mole O/1 mole CO, and 16.0 g O/1 mole O which equals 14.9 grams O.

    Good job, stoichiometry can be difficult.

  3. I am getting 7.43 grams of oxygen for part c. For the equation, I said O2 + 2C = 2CO since O can't stand alone. When i used this as the equation i got 7.43. I could be wrong but I was just wondering if this would be the answer instead.