CaCO_3 (aqueous suspension) xrightarrow{AcOH} | vedclass

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(D) The reaction sequence is as follows:$CaCO_3 (s) + 2CH_3COOH (aq) \rightarrow Ca(CH_3COO)_2 (aq) + H_2O (l) + CO_2 (g)$In the first step,$CaCO_3$ r

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$1$ and $3$ both are correct

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(D) The reaction sequence is as follows:$CaCO_3 (s) + 2CH_3COOH (aq) \rightarrow Ca(CH_3COO)_2 (aq) + H_2O (l) + CO_2 (g)$In the first step,$CaCO_3$ reacts with acetic acid $(AcOH)$ to form soluble calcium acetate,releasing $CO_2$ gas.In the second step,$Ca(CH_3COO)_2 (aq) + Na_2C_2O_4 (aq) \rightarrow CaC_2O_4 (s) + 2CH_3COONa (aq)$Calcium oxalate $(CaC_2O_4)$ is a white precipitate.Thus,$CO_2$ is evolved and both the initial substance $(CaCO_3)$ and the final precipitate $(CaC_2O_4)$ are white in colour.Therefore,both observations $1$ and $3$ are correct.

Column-$I$	Column-$II$
$(i)$ Ammonium phosphomolybdate	$(a)$ $(NH_4)_2MoO_4$
$(ii)$ Ammonium phosphate	$(b)$ $Mg^{2+} + NH_4OH$
$(iii)$ Mixture of magnesia	$(c)$ $H_3PO_4$
$(iv)$ Ammonium molybdate	$(d)$ $(NH_4)_3PO_4$
$(v)$ Phosphoric acid	$(e)$ $(NH_4)_3PO_4 \cdot 12MoO_3$

$CaCO_3$ (aqueous suspension) $\xrightarrow{AcOH}$ $\xrightarrow{Na_2C_2O_4 \text{ solution}}$
Which of the following observations is correct for the above experiment?

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$CaCO_3$ (aqueous suspension) $\xrightarrow{AcOH}$ $\xrightarrow{Na_2C_2O_4 \text{ solution}}$Which of the following observations is correct for the above experiment?