The amount of Na_2S_2O_3 cdot 5H_2O required | vedclass

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(A) The reaction between iodine and sodium thiosulfate is: $I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI$. At the equivalence point,the number of e

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$6.20$

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(A) The reaction between iodine and sodium thiosulfate is: $I_2 + 2Na_2S_2O_3 ightarrow Na_2S_4O_6 + 2NaI$. At the equivalence point,the number of equivalents of $Na_2S_2O_3 \cdot 5H_2O$ must equal the number of equivalents of $I_2$. Number of equivalents of $I_2 = 	ext{Normality} 	imes 	ext{Volume (in L)} = 0.25 \, N 	imes 0.1 \, L = 0.025 \, 	ext{eq}$. Since the $n$-factor for $Na_2S_2O_3$ in this reaction is $1$ (as $S_2O_3^{2-} ightarrow S_4O_6^{2-} + 2e^-$ per mole,but $2$ moles of thiosulfate are involved per $I_2$,the equivalent weight equals the molar mass),the number of moles of $Na_2S_2O_3 \cdot 5H_2O$ is equal to the number of equivalents. Molar mass of $Na_2S_2O_3 \cdot 5H_2O = 23 	imes 2 + 32 	imes 2 + 16 	imes 3 + 5 	imes 18 = 248 \, g/mol$. Mass required $= 	ext{Equivalents} 	imes 	ext{Equivalent weight} = 0.025 	imes 248 = 6.20 \, g$.

The amount of $Na_2S_2O_3 \cdot 5H_2O$ required to completely reduce $100 \, mL$ of $0.25 \, N$ iodine solution is $.... \, g$.

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