Compound D isNa xrightarrow{H_2O} (A) xrighta | vedclass

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(B) The reaction sequence is as follows:$1$. $Na + H_2O \rightarrow NaOH (A) + \frac{1}{2} H_2$$2$. $2NaOH (A) + CO_2 \rightarrow Na_2CO_3 (B) + H_2O$

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$Na_2S_2O_5 \cdot xH_2O$

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(B) The reaction sequence is as follows:$1$. $Na + H_2O \rightarrow NaOH (A) + \frac{1}{2} H_2$$2$. $2NaOH (A) + CO_2 \rightarrow Na_2CO_3 (B) + H_2O$$3$. $Na_2CO_3 (B) + SO_2 (excess) \rightarrow Na_2SO_3 (C) + CO_2$$4$. $Na_2SO_3 (C) + SO_2 (excess) + H_2O \rightarrow 2NaHSO_3$$5$. Upon evaporation,$2NaHSO_3$ yields $Na_2S_2O_5 (D) + H_2O$.Thus,compound $D$ is sodium metabisulfite,$Na_2S_2O_5$.

Compound $D$ is
$Na$ $\xrightarrow{H_2O} (A)$ $\xrightarrow{CO_2} (B)$ $\xrightarrow[Excess]{SO_2} (C)$ $\xrightarrow[In\ excess\ SO_2]{Evaporation} (D)$ $\xrightarrow{\Delta} (E) + f_{(g)}$
$(E) \xrightarrow{\Delta} (G) + H$

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Compound $D$ is$Na$ $\xrightarrow{H_2O} (A)$ $\xrightarrow{CO_2} (B)$ $\xrightarrow[Excess]{SO_2} (C)$ $\xrightarrow[In\ excess\ SO_2]{Evaporation} (D)$ $\xrightarrow{\Delta} (E) + f_{(g)}$$(E) \xrightarrow{\Delta} (G) + H$