Fe (finely powdered) + HCl(dil.) to P + Qupar | vedclass

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(D) The reaction between finely powdered $Fe$ and dilute $HCl$ is: $Fe + 2HCl(dil.) 	o FeCl_2 + H_2\uparrow$. Here,$P$ is $FeCl_2$ and $Q$ is $H_2$.$

Vedclass · Accepted Answer

$NH_4Cl + NH_4OH$

Vedclass · Answer

(D) The reaction between finely powdered $Fe$ and dilute $HCl$ is: $Fe + 2HCl(dil.) 	o FeCl_2 + H_2\uparrow$. Here,$P$ is $FeCl_2$ and $Q$ is $H_2$.$FeCl_2$ contains $Fe^{2+}$ ions.$1$. With $AgNO_3$: $FeCl_2 + 2AgNO_3 	o 2AgCl(s) + Fe(NO_3)_2$. $AgCl$ is a white precipitate.$2$. With $K_3[Fe(CN)_6]$: $3FeCl_2 + 2K_3[Fe(CN)_6] 	o Fe_3[Fe(CN)_6]_2 + 6KCl$. This forms a blue precipitate (Turnbull's blue).$3$. With $(NH_4)_2S$: $FeCl_2 + (NH_4)_2S 	o FeS(s) + 2NH_4Cl$. $FeS$ is a black precipitate.$4$. With $NH_4Cl + NH_4OH$: $Fe^{2+}$ ions do not form a precipitate with $NH_4OH$ in the presence of $NH_4Cl$ because the concentration of $OH^-$ ions is suppressed by the common ion effect of $NH_4^+$,which is insufficient to exceed the solubility product of $Fe(OH)_2$.

$Fe$ (finely powdered) $+ HCl(dil.) \to P + Q\uparrow$. Compound $P$ does not precipitate with:

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