8.1 g of `K_2Cr_2O_7` reacts with 12.8 g of HI according to the equation
`Cr_2O_^(2-)+HItoCrI_3+KI+I_2`
Calculate:
(a). Percentage by mass of `K_2Cr_2O_7` left unreasted.
(b). Volume of `I_2` (g) evolved, if `I_2` obtained is heated to 500 K and 1.0 atm pressure.

Explanation

Balance the redox reaction:
`14H^(o+)+cancel(6e^(-))+Cr_2O_7^(2-)to2Cr^(3+)+7H_2O`
`underline(2HItoI_2+2e^(-)+2H^(+)]xx3)`
`underline(8H^(o+)+Cr_2O_7^(2-)+6HIto2Cr^(3+)+3I_2+7H_2O)`
To balance equation for other ions, add `2K^(o+)` and `8I^(ɵ)` to both sides.
`14HI+K_2Cr_2O_7^(2-)to2CrI_3+3I_2+7H_2O`
`14xx138g of HI `requires `-=1 mol K_2Cr_2O_7-=294g`
`12.8 of HI` requires`=(294xx12.8)/(14xx128)=2.1g K_2Cr_2O_7`
(a). Weight of `K_2CrO_2O_7` unreacted `=8.1-2.1=6.0g`
`%` of `K_2Cr_2O_7` unreacted`=(6xx100)/(8.1)=74.07%`
(b). 1 " mol of "`K_2Cr_2O_7(=294g) gives =3 " mol of "I_2`
`2.1 g of K_2Cr_2O_7-=(3xx2.1)/(294)=0.021 mol I_2`
`PV=nRT`
`V_(I_2)=(nRT)/(P)=(0.021xx0.082xx500)/(1)`
`=0.861L=861 mL`