[Emma]

Hello,

I am extremely stuck with these calculations

1) From the mass of K2Cr2O7 used and the volume of solution prepared, calculate the concentration of HCrO4 - present in your stock solution.

we were not given the mass of the K2Cr2O7 , only its concentration 0.0055 and the volume of solution to be 5mL. How are we supposed to find the concentration of HCrO4 then ?

2) using 2HCrO4--+14H30+6I-->2Cr3++22H2O+3I2
I2+2S2O3--==> 2I-+S4O6--
 And the calculated molarity of HCrO4 and the volume of Na2S2O3 (i averged the two trials to be 23.15mL)
 solution required to reduce 10.00 mL of HCrO4
- solution, calculate the concentration of the Na2S2O3 solution.

so i get we are dealing with stoichiometry here , but how can I relate the number of moles of HCrO4 with the Na2S2O3 ?

Please help me its almost impossible to answer these questions.

[Archer]

we were not given the mass of the K₂Cr₂O₇ , only its concentration 0.0055 and the volume of solution to be 5mL. How are we supposed to find the concentration of

0.0055 what? I assume M but it could be ppm for all I know.

You are missing data, how much iodide did you add and at what concentration (provide units)

How many moles of HCrO₄ react with how many moles of I^- to give how many moles of I₂


How many moles of Na₂S₂O₃ react with how many moles of I₂?

Start there

[Emma]

we were not given the mass of the K₂Cr₂O₇ , only its concentration 0.0055 and the volume of solution to be 5mL. How are we supposed to find the concentration of

0.0055 what? I assume M but it could be ppm for all I know.

You are missing data, how much iodide did you add and at what concentration (provide units)

How many moles of HCrO₄ react with how many moles of I^- to give how many moles of I₂


How many moles of Na₂S₂O₃ react with how many moles of I₂?

Start there

so I finalized the number of moles of I2 to b 3.6x10^(-5) mole

i used  (0.000055 moles of HCrO4)x(3 moles of I2)/(2 moles of HCrO4) ---- using stoichiometry  and the equation

[Archer]

We can't tell you if you are right or not as you have not given the data for the iodide solution (molarity and volume)

Please provide us with all available data and full calculations (preferably formatted correctly) including all units.

We will be glad to tell you if you are right an if you're not then we can see where the errors were made.

[Archer]

I made a start on formatting your post to make it clearer to read.

Hello,

I am extremely stuck with these calculations

1) From the mass of K₂Cr₂O₇ used and the volume of solution prepared, calculate the concentration of HCrO₄^- present in your stock solution.

we were not given the mass of the K₂Cr₂O₇, only its concentration 0.0055 and the volume of solution to be 5mL. How are we supposed to find the concentration of HCrO₄then ?

2) using 2HCrO₄ +14H₃0+6I^-+2Cr³⁺  22H₂O+3I₂
I₂+2S₂O₃^2- 2I^-+S₄O₆^2-
 And the calculated molarity of HCrO₄ and the volume of Na₄S₂O₃ (i averged the two trials to be 23.15mL)
 solution required to reduce 10.00 mL of HCrO₄
- solution, calculate the concentration of the Na₂S₂O₃  solution.

so i get we are dealing with stoichiometry here , but how can I relate the number of moles of HCrO₄ with the Na₂S₂O₃?

Please help me its almost impossible to answer these questions.

I know it's time consuming but it makes our job much easier, it takes longer doing it retrospectivly as I have just discovered. Probably introduced a mistake it two as I am on the smallest HTC screen ever made