Question

Iron in the 2 oxidation state reacts with potassium dichromate to produce Fe3 and Cr3 according to the equation: 6 Fe2 (aq) Cr2O72-(aq) 14 H (aq) <----> 6 Fe3 (aq) 2 Cr3 (aq) 7 H2O(l) How many milliliters of 0.2937 M K2Cr2O7 are required to titrate 132.0 mL of 0.1782 M Fe2 solution

Answer 1

Given :

Volume of $Fe^{2+}$ , V = 132 mL .

Molarity of $Fe^{2+}$, M = 0.1782 M .

To Find :

How many milliliters of 0.2937 M $K_2Cr_2O_7$ are required to titrate 132.0 mL of 0.1782 M $Fe^{2+}$ solution .

Solution :

Moles of  $Fe_2$ :

$n=0.1782\times \dfrac{132}{1000}\ mol\\\\n=0.264\ mol$

Now , 1 mole of $K_2Cr_2O_7$ reacts with 6 mole of $Fe^{2+}$ .

So , moles of $K_2Cr_2O_7$ required is :

$N=\dfrac{0.264}{6}=0.044\ mol$

Volume required is :

$V=\dfrac{N}{M}\\\\V=\dfrac{0.044}{0.2937}\ L\\\\V=0.15\ L=150\ mL$

Hence , this is the required solution .