Blog #4

Questions regarding Gen Chem 2 class:

1) Most valuable lesson?

-The most valuable lesson while in this class, really not only applies to this class, but to them all. However this class really made me understand that it is very important to stay on top of your work and to not wait until the night before to study. It is much better to spend time looking at smaller portions of a chapter(s) over a week or so, than trying to look over everything in one night. 


2) Most challenging concept?

-The most challenging concept thus far has definitely been dealing with and balancing the oxidation reduction equations. I really struggle with grasping the whole concept of it all, and never really was able to confidently answer a question regarding this material.

3) Advice for future Gen Chem 2 students?

-For future Gen Chem 2 students, I would advise that you definitely read the material you are covering in class, before that class. As well as working the in chapter problems and end of chapter problems as you go along, being sure to completely understand any concepts you are struggling with. Also would recommend learning how to do every single sapling before the exams, because if you can do all the saplings without looking at the book, you can make A's on all the test. No questions about it. Lastly, do not do not do not wait until the last minute to study or do anything for that matter. May work in other classes, but not this one.

Blog #3: Question Response

Question asked by Maggie Calvert:

Sodium oxalate, Na2C2O4, in solution is oxidized to CO2(g) by MnO4- which is reduced to Mn2+. A 50.1ml volume of a solution of MnO4- is required to titrate a 0.339g sample of sodium oxalate. This solution of MnO4- is used to analyze uranium-containing samples. A 4.62g sample of a uranium-containing material requires 32.5ml of the solution for titration. The oxidation of the uranium can be represented by the change UO2+-->UO2 2+. Calculate the percentage of uranium in the sample.

Solution:

First the entire reaction was wrote out as
2 KMnO4 + 5 Na2C2O4 + 8 H2SO4 ---> 2 MnSO4 + K2SO4 + 5 Na2SO4 + 10 CO2 + 8 H2O
The molar mass of sodium oxalate and potassium permanganate were calculated:

Molar mass of Na2C2O4 is 134 g/mol

Molar mass of KMnO4 is 158 g/mol 

 

Using the reaction above, it can be seen that

2 moles of KMnO4 reacts with  5 *134 of Na2C2O4  

X moles of KMnO4 reacts with  5*0.339 g of Na2C2O4  

X = ( 2*5*0.339g) / (5*134g)

   = .00056 moles of KMnO4 

 

So Molarity of MnO4 - , M = no.of moles / Volume of solution in L

                                           = .00056 moles / 0.0501 L

                                           = 0.01108 M 

 

The reaction of uranium sulfate with potassium permanganate with water is wrote as:

5 U(SO4)2 + 2 KMnO4 + 2 H2O ----> 2 H2SO4 + K2SO4 + 2 MnSO4 + 5 UO2SO4

Volume of MnO4- required for this titration is , V = 32.5 mL = 0.0325 L

So no.of moles of MnO4- , n = Molarity * Volume in L

                                             = 0.01108 M * 0.0325 L

                                             = .00036  moles of MnO4- 

 

Using the equation directly above, 

5 moles of U(SO4)2  reacts with 2 moles of MnO4 -

So Y moles of U(SO4)2  reacts with .00036 moles of MnO4-

Y = ( 5*.00036 moles ) / 2

   = .0009 moles

Molar mass of U(SO4)2  = 430 g

So mass of U(SO4)2  is , m = No.of moles * Molar mass = .3876 g 

Therefore the percentage of Uranium in the sample is calculated by

( .3876g/ 4.62g) *100 = 8.39 % 

 

Insightful Response:

This was a very good and hard (haha) question as it utilized being able to write the proper equations and balance them for the reactions, using the info learned from this chapter, as well as incorporating things we learned previously in chem.

Sources:

 

Chemistry Book, Chapter 18 Electrochemistryfor being able to write out the reactions

Using previous knowledge of titrations and basic chemistry for mathematical calculations.