$$4.9$$ g of $$K2Cr2O7$$ is taken to prepare $$0.1$$ L of the solution $$10$$ mL of this solution is further taken to oxidize $$Sn2+$$ ion into $$Sn4+$$ ion $$Sn4+$$ so produced is used in second reaction to prepare $$Fe3+$$ ion then the millimoles of $$Fe3+$$ ion formed will be (assume$$ all other components are in sufficient $$amount) [Molar mass of $$K2Cr2O7$$ $$=$$ $$294$$ g].

Question

$$4.9$$ g of $$K2Cr2O7$$ is taken to prepare $$0.1$$ L of the solution $$10$$ mL of this solution is further taken to oxidize $$Sn2+$$ ion into $$Sn4+$$ ion $$Sn4+$$ so produced is used in second reaction to prepare $$Fe3+$$ ion then the millimoles of $$Fe3+$$ ion formed will be (assume$$ all other components are in sufficient $$amount) [Molar mass of $$K2Cr2O7$$ $$=$$ $$294$$ g].

Infinity Learn · Accepted Answer

$$K2Cr2O7$$ $$+$$ $$Sn2+$$ →  $$Sn4+$$ $$+$$ $$Cr3+Sn4+$$ $$+$$ $$Fe2+$$→ $$Fe3+Meq$$. of $$Sn4+$$ $$=$$ Meq of  $$K2Cr2O7Meq$$. of $$Sn4+$$ $$=$$ Meq of $$Fe3+Meq$$. of $$Fe3+$$ $$=$$ meq of $$K2Cr2O7NK2Cr2O7$$ $$=$$ $$4.9$$ × $$6294$$ × $$0.1$$ $$=$$ $$1Meq$$. of $$Fe+3$$ $$=(NV)$$ of $$K2Cr2O7millimol$$ x $$n-factor$$ $$=$$ $$1$$ x $$10millimol$$ $$=$$ $$10$$

Detailed Solution

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