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How many ml of acidified $\ce{K2Cr2O7}$ solution, which includes 21.6 g of dichromate ions per Litre, are required for the oxidation of 44.4 g of $\ce{CH3CH2CH(OH)CH3}$?

Obviously 44.4 g of $\ce{CH3CH2CH(OH)CH3}$ means 0.6 moles. So from the stoichiometry of the the oxidation we find out that 0.2 moles of $\ce{K2Cr2O7}$ are needed. I have trouble finding the concentration of $\ce{K2Cr2O7}$ though and thats why I need help.

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First, you need a balanced equation for the oxidation. The gives you the mole ratio of dichromate to substrate. You know how many moles of substrate you have, you know how many moles of dichromate you need. If you have 21.6 g of dichromate ions (as such)/liter, then you know how many moles/liter you have. You know the number of moles you need, you know the number of moles/liter you have, you know the number of liters you need.

You will be doing your Jones oxidation in acid, of course. Collins reagent is a sweeter deal for workup, since MEK is 27.5 g/100 ml water-soluble.

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To calculate the dichromate concentration (in moles/l) of the solution, simply divide the given mass of dichromate per litre (21,6 g) by the molar mass of $\ce{Cr2O7^{2-}}$.

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