Question

How many mL of 0.125 M Cr3+ must be reacted with 12 mL of 0.2 M MnO4 -, if the redox products are CrO7 2- and Mn2+ ?

Answer 1

Answer : 19.2 ml of $Cr^{+3}$ reacted with $MnO^-_4$.

Solution : Given,

Molarity of $Cr^{+3}$ = 0.125 mole/L

Molarity of $MnO^-_4$ = 0.2 mole/L

Volume of $MnO^-_4$ = 12 ml = 0.012 L     (1 L = 1000 ml)

The given chemical redox reaction is,

$Cr^{3+}+MnO^-_4\rightarrow CrO^{2-}_4+Mn^{2+}$

From the given reaction, we conclude that the 1 mole of $Cr^{+3}$ react with the 1 mole of $MnO^-_4$.

Formula used : $M_1V_1=M_2V_2$

where,

$M_1$ = Molarity of $Cr^{+3}$

$V_1$ = Volume of $Cr^{+3}$

$M_2$ = Molarity of $MnO^-_4$

$V_2$ = Volume of $MnO^-_4$

Now put all the given values in above formula, we get the volume of $Cr^{+3}$.

$0.125mole/L\times V_1=0.2mole/L\times 0.012L\\V_1=0.0192L=19.2ml$

Therefore, 19.2 ml of $Cr^{+3}$ reacted with $MnO^-_4$.