Help with balancing equation and concentration.

[Trufflehog]

Hi

 

I asked the question below on another website and was told that the answer is 66.66666667ml not 600ml. Any chance of a second opinion? The answer in my worksheet states it's 600mL

 

 

I was just wondering if someone can check over my working for this problem and let me know if I went wrong anywhere or any useful tips that they wish to share?

(Everything that I wrote is in the *** triple stars ***)

Dichromate ions, Cr2O7^2- reacts with iodide ions I^‐,in acidic aqueous solution to form Cr³⁺ ions, iodine molecules I₂ and water, according to the following unbalanced equation:

Cr₂O₇^2- + H⁺ + I^- → Cr³⁺ + H₂O + I₂

***Cr₂O₇^2- + 14H⁺ + 2I^- → 2Cr³⁺ + 7H₂O + I₂***

A solution of 100 mL of 0.1M K₂Cr₂O₇ in dilute acid is prepared. How many mL of a 0.1M aqueous solution of potassium iodide (KI) would be required for complete reaction?

***I use HCl as the dilute acid***

***K₂Cr₂O₇ + 14HCl + 2KI → 2CrCl₃ + 7H₂O + I₂ + KCl***

***Balance this equation***

***K₂Cr₂O₇ + 14HCl + 6KI → 2CrCl₃ + 7H₂O + 3I₂ + 8KCl***

***0.1M K₂Cr₂O₇/0.1L = 0.01mol***

***K₂Cr₂O₇ 0.01mol * (KI 6 mol)/(K₂Cr₂O₇ 1 mol) = 0.06mol***

***0.06mol/V = 0.1M → V = 0.06mol/ 0.1M = 0.6L***

***Therefore 600mL of 0.1M aqueous solution of potassium iodide (KI) is needed for a complete the reaction***

[hypervalent_iodine]

I haven't checked the balancing, but assuming it's correct then your calculation for volume is correct and 600mL would be the answer.

[Fuzzwood]

The balancing is done incorrectly. Dichromate is reduced to 2 Cr(III) ions, taking up a total of 6 electrons in the progress. Your reaction only deals with 2 of those. Although further ahead you do manage to come up with the correct equation.

Edited June 11, 2014 by Fuzzwood

[Trufflehog]

Oh, I see what I did there. I should have had Cr₂O₇^2- + 14H⁺ + 6I^- → 2Cr³⁺ + 7H₂O + 3I₂. Wow that would have saved the strange road where I got the answer from. Thanks for pointing that out and confirming the answer.