Calculating the Number of Grams of Ethanedioate Ion in 1 dm3

[crays]

Hi, i wonder how could i calculate the number of grams of ethanedioate ion, C₂O₄^2- in one dm³

Given:
FA1 was made by dissolving 6 g of metal ethanedioate, MC₂O₄ is dissolved in dilute sulpuhric acid and making up to 1 dm³ with pure water.

FA2 contains 2.38g of manganate (VII) ion per dm³.

25.0 cm³ of FA1 requires 19.65 cm³ of FA2 for complete reaction.

As I'm pretty weak in chemistry, i tried my best to solve the question. I used the (M_AV_A)/(M_BV_B) = ratio/ratio method.
But not sure if I'm right.

I also cannot differentiate if its suppose to be concentration or just pure mass. can someone help me here?

FA2 contains 2.38g of manganate (VII) ion per dm³.

does it means it have 2.38g (mass) or 2.38gdm^-3 (concentration) ?

 

[Borek]

First of all - you need balanced equation of reaction between oxalate (ethanodioate) and permanganate (manganate(VII)).

Amount of permanganate in FA2 is given as mass per volume, so it is concentration.

 

[crays]

Erm, but it wanted ions. So i use half equation?

2MnO₄^- + 5C²O₄^2- + 16H⁺ ----> 2Mn²⁺ + 10CO₂ + 8H₂O

Molarity of FA2 = 2.38/119.4

Molarity of C²O₄^2- x 25.0 divided by 2.38/119.4 x 19.65 = 5/2

Correct ?

 

[Borek]

Erm, but it wanted ions. So i use half equation?

2MnO₄^- + 5C²O₄^2- + 16H⁺ ----> 2Mn²⁺ + 10CO₂ + 8H₂O

That's a correct equation. Not a half equation, but a full one.

Molarity of FA2 = 2.38/119.4

That's correct, although just dividing you get number of moles. However, mass was given per L, so it turns out the same.

Molarity of C²O₄^2- x 25.0 divided by 2.38/119.4 x 19.65 = 5/2

You have lost me here, although it is very likely that you are OK. Just post the final number.