Stoichiometry Did I work this problem correctly?

[bedizzy]

Stoichiometry... Did I work this problem correctly?

1. We dissolve 3.778 grams of a sample that contains some sodium oxalate, Na2C2O4, in water and acidify the solution with excess sulfuric acid. The sample requires 18.74 mL of 0.08395 M KMnO4, potassium permanganate, for complete reaction according to the reaction below. What is the percent Na2C2O4 in the sample? Assume that no other component reacts with the potassium permanganate.

8H2SO4 + 2KMnO4 + 5Na2C2O4 --> 8H2O + 2MnSO4 + 10CO2 + 5Na2SO4 + K2SO4



My solution:

.01874 L * 0.08395 M KMnO4 = 0.001573223 mols KMnO4

0.001573223 * (5 mols Na2C2O4/2 mols KMnO4)
= .0039330575 mols Na2C2O4

.0039330575 * (116.04 g) * (100/3.778 g) = 12.08 %


(1 mol Na2C2O4 = 116.04 g)



I didn't round any numbers until the final answer. Does this look correct? If not where do you see an error? Thanks in advance for checking...

 

[bedizzy]

And if it's not too much can you check this one too:1. How many mL of water must be added to 5.00 mL of 12.0 M HCl to prepare a hydrochloric acid solution that is 0.600 M? Assume that the volumes are additive.My solution:

(.00500 L * 12.0 M HCl)/(0.600 M) = .1 L = 100. mL

 

[symbolipoint]

Your steps are good in the oxalate analysis.

 

[symbolipoint]

You only are trying to make a simple dilution for the HCl 12M to 0.600M exercise. use a variable to represent the volume of water to add. You already know that the MOLES of HCl will not change. Next you want to try to create the equation ... You will need to use that variable somewhere. The variable is to be used in a denominator. Why not in a numerator?

 

[symbolipoint]

I did not actually check your result in the HCl example; I only vaguely described what you can do to solve the problem. I like to arrange an equation using a variable for problems like that one.

 

[bedizzy]

thanks for the feedback, this is my rationale for setting up the problem the way I did:


V₁M₁ = V₂M₂

(.00500 * 12.0 M HCl) = (.600 M * V₂)

divide both sides by .600 M

V₂ = (.00500 * 12. M HCl)/(.600 M)

V₂ = 100. mL

 

[symbolipoint]

thanks for the feedback, this is my rationale for setting up the problem the way I did:


V₁M₁ = V₂M₂

(.00500 * 12.0 M HCl) = (.600 M * V₂)

divide both sides by .600 M

V₂ = (.00500 * 12. M HCl)/(.600 M)

V₂ = 100. mL

Yes, that is a good way to do it. It is not like my "use variable for volume to add" suggestion, but your method looks good. Notice your final volume will be 100 ml. So, how much volume was added to the initial 5 ml. of solution?