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  4. Volume of 0.1 M K2Cr2O7 required to oxidise 35 mL of 0.5 M FeSO4 solution is

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Q. Volume of 0.1 M $K_2Cr_2O_7$ required to oxidise 35 mL of 0.5 M $FeSO_4$ solution is

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Solution:

$K_{2}Cr_{2}O_{7}+4H_{2}SO_{4} \rightarrow K_{2}SO_{4}+Cr_{2}\left(SO_{4}\right)_{3}+4H_{2}O+3\left[O\right]$
$2FeSO_{4}+H_{2}SO_{4}+\left[O\right] \rightarrow Fe_{2}\left(SO_{4}\right)_{3}+H_{2}O \times3$
$\frac{K_{2}Cr_{2}O_{7}+7 H_{2}SO_{4}+6FeSO_{4} \rightarrow K_{2}SO_{4}+Cr_{2} \left(SO_{4}\right)_{3} +7H_{2}O+3Fe_{2}\left(SO_{4}\right)_{3}}{\frac{M_{1} V_{1}}{n_{1}}=\frac{M_{2}V_{2}}{n_{2}}}$
$(K_{2}Cr_{2}O_{7}) (FeSO_{4})$
$\frac{0.1\times V_{1}}{1}=\frac{0.5\times35}{6\times0.1}$
$V_{1}=\frac{0.5\times35}{6\times0.1}=29.2\,mL$