How many milliliters of 0.025 M `K_2Cr_2O_7` are required to react completely with 25.0 mK of 0.20 M solution of `FeSO_4` ?

Explanation

The balanced chemical equation is
`3Fe^(2+)+Cr_2O_7^(2-)+14H^(+)rarr6Fe^(3+)+2Cr^(3+)+7H_2O`
1 mol of `K_2Cr_2O_7 =6 " mol of "FeSO_4`
Moles of `FeSO_4` present in 25.0 mL of 0.20 M solution
`(0.20)/1000xx25.0 =0.005 mol`
Now 6 mol of `FeSO_4` require `K_2Cr_2O_7`
= 1mol
`=1/6xx0.005 = 0.000833 ` mol
Let us calculate the volume of `K_2Cr_2O_7` solution containing 0.000833 mol
0.025 mol of `K_2Cr_2O_7` are present in
=100 mL
0.000833 mol of `K_2Cr_2O_7` are present in
`=(1000xx0.000833)/(0.025)=33.3mL`