Consider the following reaction

 

As shown, A, B, C, and D are each in turn.

The reaction between sodium chloride and sulfuric acid occurs as follows:
$\mathit{NaCl}\left(s\right)+H_2{\mathit{SO}}_4\left(\mathit{aq}\right)\to {\mathit{Na}}_2{\mathit{SO}}_4\left(\mathit{aq}\right)+\mathit{HCl}\left(\mathit{aq}\right)$

The reaction between hydrogen chloride and iron is as follows:
$\mathit{HCl}\left(\mathit{aq}\right)+\mathit{Fe}\left(s\right)\to \mathit{Fe}{\mathit{Cl}}_2\left(\mathit{aq}\right)+H_2\left(g\right)\uparrow$

The reaction between hydrogen chloride with ammonia gas is as follows:
$\mathit{HCl}\left(\mathit{aq}\right)+{\mathit{NH}}_3\left(g\right)\to {\mathit{NH}}_4\mathit{Cl}\left(s\right)$ 

The reaction between hydrogen chloride and Lead nitrate is as follows: $\mathit{HCl}\left(\mathit{aq}\right)+{\mathit{Pb}\left({\mathit{NO}}_3\right)}_2\left(\mathit{aq}\right)\to {\mathit{PbCl}}_2\left(s\right)+{\mathit{HNO}}_3\left(\mathit{aq}\right)$
 

Hence, in the given reaction, here A, B, C, and D respectively are A = conc. $H_2{\mathit{SO}}_4$; B = Fe; C = ${\mathit{NH}}_3$; D = ${\mathit{Pb}\left({\mathit{NO}}_3\right)}_2$.