How many mL of 0.125 M Cr^(3+)must be rected with 12.00 mL of 0.200 M MnO_(4)^(-) if theredox products are Cr_(2)O_(7)^(2-) andMn^(2+) ?

How many mL of 0.125 M Cr^(3+)must be rected with 12.00 mL of 0.200 M

1.	How many mL of 0.125 M Cr^(3+)must be rected with 12.00 mL of 0.200 M MnO_(4)^(-) if theredox products are Cr_(2)O_(7)^(2-) andMn^(2+) ?
Answer» 8 ML 16 mL 24 mL 32 mL Solution :Reduction equation OXIDATION equation `2Cr^(3+)+7_(2)OrarrCr_(2)O_(7)^(2-)+14H^(+)+6e^(-)xx5` balanced redox equation applying molarity equation `therefore (0.125xxV)/(10)=(0.2000xx12.00)/(6)` or `V_(1)=(0.200xx12.00)/(6)xx(10)/(0.125)=32 mL`