Formic acid is heated with conc. H_2SO_4 at 1 | vedclass

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Question

(A) $HCOOH \xrightarrow{	ext{Conc. } H_2SO_4} \underset{(B)}{CO} + \underset{(A)}{H_2O}$ $3 \underset{(B)}{CO} + Fe_2O_3 \xrightarrow{\Delta} \unders

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$Fe$

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(A) $HCOOH \xrightarrow{	ext{Conc. } H_2SO_4} \underset{(B)}{CO} + \underset{(A)}{H_2O}$ $3 \underset{(B)}{CO} + Fe_2O_3 \xrightarrow{\Delta} \underset{(D)}{2Fe} + \underset{(C)}{3CO_2}$ $CaCO_3 + 2HCl 	ext{ (dil.)} \longrightarrow \underset{(C)}{CO_2} + H_2O + CaCl_2$ From the reactions,$B$ is $CO$ and $C$ is $CO_2$. The reduction of $Fe_2O_3$ by $CO$ yields $Fe$ $(D)$ and $CO_2$ $(C)$. Therefore,$D$ is $Fe$.

List-$I$ (Hydrides)	List-$II$ (Nature)
$a. MgH_2$	$i. \text{Electron precise}$
$b. GeH_4$	$ii. \text{Electron deficient}$
$c. B_2H_6$	$iii. \text{Electron rich}$
$d. HF$	$iv. \text{Ionic}$

Formic acid is heated with conc. $H_2SO_4$ at $100^{\circ}C$ to form $A$ and $B$. When $Fe_2O_3$ is heated strongly with $B$,the products formed are $C$ and $D$. $C$ can also be obtained by reacting $CaCO_3$ with dil. $HCl$. What is $D$?

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Match List-$I$ with List-$II$.

List-$I$ (Hydrides) List-$II$ (Nature)

$a. MgH_2$ $i. \text{Electron precise}$

$b. GeH_4$ $ii. \text{Electron deficient}$

$c. B_2H_6$ $iii. \text{Electron rich}$

$d. HF$ $iv. \text{Ionic}$

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The incorrect statement among the following is:

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