6 times 10^{-3} text{ mole} of K_2Cr_2O_7 rea | vedclass

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(A) The reaction involves the oxidation of $X^{n+}$ to $XO_3^-$ and the reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$.According to the law of equivalence,t

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(A) The reaction involves the oxidation of $X^{n+}$ to $XO_3^-$ and the reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$.According to the law of equivalence,the number of gram equivalents of the oxidizing agent equals the number of gram equivalents of the reducing agent.$	ext{Equivalents of } K_2Cr_2O_7 = 	ext{Equivalents of } X^{n+}$$(	ext{n-factor})_{K_2Cr_2O_7} 	imes 	ext{moles of } K_2Cr_2O_7 = (	ext{n-factor})_{X^{n+}} 	imes 	ext{moles of } X^{n+}$For $K_2Cr_2O_7$,the change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are $2$ atoms of $Cr$,the n-factor is $2 	imes (6 - 3) = 6$.For $X^{n+}$,the oxidation state changes from $+n$ to $+5$ (in $XO_3^-$). The n-factor is $(5 - n)$.Substituting the values: $6 	imes (6 	imes 10^{-3}) = (5 - n) 	imes (9 	imes 10^{-3})$$36 = (5 - n) 	imes 9$$4 = 5 - n$$n = 1$.

$6 \times 10^{-3} \text{ mole}$ of $K_2Cr_2O_7$ reacts completely with $9 \times 10^{-3} \text{ mole}$ of $X^{n+}$ to give $XO_3^-$ and $Cr^{3+}$. The value of $n$ is:

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