Question fbe82

Aug 18, 2015

The mixture is 56.33% ${\text{KMnO}}_{4}$ and 43.67% ${\text{K"_2"Cr"_2"O}}_{7}$.

Explanation:

The keys to solving this problem are the balanced chemical equations for the three redox reactions that take place.

Now, I assume that you are supposed to write and balance these equations, but I cannot do that here because that would make for a very long answer.

I wil lalso skip any explanation as to why the reactions produce what they produce and how the redox titration between iodine and sodium thiosulfate works.

So, here's what happens. In acidic medium, which in this case usually means in the presence of sulfuric acid, potassium iodide, the iodide ion, ${\text{I}}^{-}$, will be oxidized to iodine, ${\text{I}}_{2}$, according to these balanced chemical equations

$2 K M n {O}_{4} + 10 K I + 8 {H}_{2} S {O}_{4} \to 2 M n S {O}_{4} + 6 {K}_{2} S {O}_{4} + 5 {I}_{2} + 8 {H}_{2} O$

and

${K}_{2} C {r}_{2} {O}_{7} + 6 K I + 3 {H}_{2} S {O}_{4} \to C {r}_{2} {\left(S {O}_{4}\right)}_{3} + 3 {I}_{2} + 4 {K}_{2} S {O}_{4} + 7 {H}_{2} O$

The impoprtant things to notice here are the mole ratios that exist between potassium permanganate and iodine, in the first reaction, and potassium dichromate and iodine in the second reaction.

More specifically, you have a $2 : 5$ mole ratio between ${\text{KMnO}}_{4}$ and ${\text{I}}_{2}$ on one hand, and a $1 : 3$ mole ratio between ${\text{K"_2"Cr"_2"O}}_{7}$ and ${\text{I}}_{2}$ on the other. Keep this in mind.

Now focus on the reaction between iodine and sodium thiosulfate, which looks like this - I'll show you the net ionic equation

${I}_{2} + 2 {S}_{2} {O}_{3}^{2 -} \to {S}_{4} {O}_{6}^{2 -} + 2 {I}^{-}$

Notice that you need 2 moles of sodium thiosulfate in order to reduce one mole of iodine. This means that you can use the molarity of the sodium thiosulfate solution to find the moles of iodine present

$C = \frac{n}{V} \implies n = C \cdot V$

$n = \text{0.15 M" * 100 * 10^(-3)"L" = "0.015 moles}$

This means that the number of moles of iodine was

0.015color(red)(cancel(color(black)("moles"Na_2S_2O_3))) * ("1 moles I"""_2)/(2color(red)(cancel(color(black)("moles"Na_2S_2O_3)))) = "0.0075 moles I"""_2

Now all you have to do is use basic algebra to write a system of equations using the masses of potassium permanganate, which we'll call $x$, and potassium dichromate, which we'll call $y$.

You know that the two masses add up to give the total mass of the sample, so your first equation is

$x + y = \text{0.561 g}$

Now focus on the number of moles of iodine. The total number of moles of iodine is 0.0075. Use the aforementioned mole ratios and the molar masses of the two compounds to write

underbrace(5/2 * x/("158.03 g/mol"))_(color(blue)("moles of I"""_2" from "KMnO_4)) + underbrace(3 * y/("254.18 g/mol"))_(color(red)("moles of I"""_2" from "K_2Cr_2O_7)) = "0.0075 moles"

Here's what that equation tells you. $\frac{x}{\text{158.03 g/mol}}$ represents the number of moles of potassium permanganate present in solution. If you take into account the fact that every 2 moles of ${\text{KMnO}}_{4}$ produce 5 moles of iodine, you have $\frac{5}{2} \cdot \frac{x}{\text{158.03 g/mol}}$ representing the number of moles of iodine produced be potassium permanganate.

The same is true for potassium dichromate. Solve this simple system of equations to get the masses you're interested in

$x = 0.561 - y$

$\frac{5}{2} \cdot \frac{0.561 - y}{158.03} + 3 \cdot \frac{y}{254.18} = 0.0075$

This is equivalent to

$412.59 - 735.45 y + 474.09 y = 348.67$

$- 261.36 y = - 63.92 \implies y = \frac{\left(- 63.92\right)}{\left(- 261.36\right)} = \textcolor{g r e e n}{\text{0.245 g}}$

This means that $x$ will be

$x = 0.561 - 0.245 = \textcolor{g r e e n}{\text{0.316 g}}$

So, the mixture contained 0.316 g of potassium permanganate and 0.245 g of potassium dichromate. This means that the percent composition of the mixture was

(0.316color(red)(cancel(color(black)("g"))))/(0.561color(red)(cancel(color(black)("g")))) * 100 = color(green)(56.33%"KMnO"_4)

and

(0.245color(red)(cancel(color(black)("g"))))/(0.561color(red)(cancel(color(black)("g")))) * 100 = color(green)(43.67%"K"_2"Cr"_2"O"_7)#