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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.

Answer» <html><body><p></p>Solution :Balance the redox reaction: <br/> `<a href="https://interviewquestions.tuteehub.com/tag/14h-273936" style="font-weight:bold;" target="_blank" title="Click to know more about 14H">14H</a>^(o+)+cancel(6e^(-))+Cr_2O_7^(2-)to2Cr^(3+)+7H_2O` <br/> `underline(2HItoI_2+2e^(-)+2H^(+)]xx3)` <br/> `underline(8H^(o+)+Cr_2O_7^(2-)+6HIto2Cr^(3+)+3I_2+7H_2O)` <br/> To balance equation for other ions, add `2K^(o+)` and `8I^(ɵ)` to both sides. <br/> `14HI+K_2Cr_2O_7^(2-)to2CrI_3+3I_2+7H_2O` <br/> `14xx138g of <a href="https://interviewquestions.tuteehub.com/tag/hi-479908" style="font-weight:bold;" target="_blank" title="Click to know more about HI">HI</a> `requires `-=1 mol K_2Cr_2O_7-=294g` <br/> `12.8 of HI` requires`=(294xx12.8)/(14xx128)=2.1g K_2Cr_2O_7` <br/> (a). <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of `K_2CrO_2O_7` unreacted `=8.1-2.1=6.0g` <br/> `%` of `K_2Cr_2O_7` unreacted`=(6xx100)/(8.1)=74.07%` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>). 1 " mol of "`K_2Cr_2O_7(=294g) gives =3 " mol of "I_2` <br/> `2.1 g of K_2Cr_2O_7-=(3xx2.1)/(294)=0.021 mol I_2` <br/> `PV=nRT` <br/> `V_(I_2)=(nRT)/(P)=(0.021xx0.082xx500)/(1)` <br/> `=0.861L=861 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>`</body></html>


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