Salt X is:X text{ (Colourless salt)} xrightar | vedclass

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(B) $1$. The reaction $X$ $\xrightarrow[\Delta]{NaOH} Y$ $\xrightarrow{HCl} 	ext{White fumes}$ indicates that $Y$ is ammonia $(NH_3)$ gas,which react

Vedclass · Accepted Answer

$(NH_4)_2C_2O_4$

Vedclass · Answer

(B) $1$. The reaction $X$ $\xrightarrow[\Delta]{NaOH} Y$ $\xrightarrow{HCl} 	ext{White fumes}$ indicates that $Y$ is ammonia $(NH_3)$ gas,which reacts with $HCl$ to form $NH_4Cl$ (white fumes). This confirms that $X$ contains the ammonium ion $(NH_4^+)$.$2$. The reaction $X$ $\xrightarrow{CaCl_2} Z 	ext{ (White ppts.)}$ $\xrightarrow{KMnO_4} 	ext{Colourless solution}$ indicates that $Z$ is calcium oxalate $(CaC_2O_4)$,which is a white precipitate. Oxalate ions $(C_2O_4^{2-})$ react with $KMnO_4$ in an acidic medium to form a colourless solution ($CO_2$ and $Mn^{2+}$).$3$. Combining these,the salt $X$ is ammonium oxalate,$(NH_4)_2C_2O_4$.

Salt $X$ is:
$X \text{ (Colourless salt)}$ $\xrightarrow[\Delta]{NaOH} Y$ $\xrightarrow{HCl} \text{White fumes}$
$X \text{ (Colourless salt)}$ $\xrightarrow{CaCl_2} Z \text{ (White ppts.)}$ $\xrightarrow{KMnO_4} \text{Colourless solution}$

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Salt $X$ is:$X \text{ (Colourless salt)}$ $\xrightarrow[\Delta]{NaOH} Y$ $\xrightarrow{HCl} \text{White fumes}$$X \text{ (Colourless salt)}$ $\xrightarrow{CaCl_2} Z \text{ (White ppts.)}$ $\xrightarrow{KMnO_4} \text{Colourless solution}$