Question

How many ml of 0.125 m cr3+ must be reacted with 12.00 ml of 0.200 m mno4- if the redox products are cro72- and mn2+?

Answer 1

Answer:- 32.0 mL

Solution:- The balanced equation for the redox reaction is:

$6MnO_4^-+10Cr^+^3+11H_2O\rightarrow 6Mn^+^2+5Cr_2O_7^-^2+22H^+$

From balanced equation, there is 6:10 mol ratio between $MnO_4^-$ and $Cr^+^3$ . From give volume and molarity, moles of $MnO_4^-$ are calculated and then using mol ratio the moles of chromium ion are calculated. Further, on dividing the moles by molarity to get the volume and the calculations are shown by using dimensional analysis as:

$12.00mL(\frac{1L}{1000mL})(\frac{0.200molMnO_4^-}{1L})(\frac{10molCr^+^3}{6molMnO_4^-})(\frac{1L}{0.125mol})(\frac{1000mL}{1L})$

= $32.0mLCr^+^3$

So, 32.0 mL of $Cr^+^3$ are required to complete the reaction.