Metallic tin in the presence of HCl is oxidiz | vedclass

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(A) The chemical reaction is: $3Sn + K_2Cr_2O_7 + 14HCl \rightarrow 3SnCl_4 + 2CrCl_3 + 2KCl + 7H_2O$. In this reaction,$Sn$ is oxidized from $Sn^0$ t

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$168.49$

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(A) The chemical reaction is: $3Sn + K_2Cr_2O_7 + 14HCl ightarrow 3SnCl_4 + 2CrCl_3 + 2KCl + 7H_2O$. In this reaction,$Sn$ is oxidized from $Sn^0$ to $Sn^{4+}$,so the n-factor for $Sn$ is $4$. Equivalents of $Sn = \frac{	ext{mass}}{	ext{equivalent weight}} = \frac{1}{118.7/4} = \frac{4}{118.7} \approx 0.0337 \ eq$. For a decinormal solution,normality $N = 0.1 \ N$. Equivalents of $K_2Cr_2O_7 = N 	imes V(L) = 0.1 	imes V$. Since equivalents of $Sn$ must equal equivalents of $K_2Cr_2O_7$: $0.1 	imes V = 0.0337$. $V = 0.337 \ L = 337 \ mL$. However,using the standard stoichiometry where $3 \ mol$ of $Sn$ reacts with $1 \ mol$ of $K_2Cr_2O_7$ $(6 \ eq)$: $1 \ g$ of $Sn = \frac{1}{118.7} \ mol$. $mol$ of $K_2Cr_2O_7 = \frac{1}{3} 	imes \frac{1}{118.7} \ mol$. Equivalents of $K_2Cr_2O_7 = 6 	imes \frac{1}{3 	imes 118.7} = \frac{2}{118.7} \approx 0.01685 \ eq$. $0.1 	imes V(L) = 0.01685 \implies V = 0.1685 \ L = 168.5 \ mL$.

Metallic tin in the presence of $HCl$ is oxidized by $K_2Cr_2O_7$ to stannic chloride. What volume of decinormal dichromate solution would be reduced by $1 \ g$ of tin? (Atomic weight of $Sn = 118.7 \ g/mol$)

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