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(A) When mercuric chloride $(HgCl_2)$ reacts with stannous chloride $(SnCl_2)$,it is first reduced to mercurous chloride $(Hg_2Cl_2)$:$2HgCl_2 + SnCl_

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liquid $Hg$ and $SnCl_4$

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(A) When mercuric chloride $(HgCl_2)$ reacts with stannous chloride $(SnCl_2)$,it is first reduced to mercurous chloride $(Hg_2Cl_2)$:$2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$When excess stannous chloride is added,the mercurous chloride is further reduced to metallic mercury $(Hg)$:$Hg_2Cl_2 + SnCl_2 \rightarrow 2Hg(l) + SnCl_4$Thus,the final products obtained with excess stannous chloride are liquid $Hg$ and $SnCl_4$.

When mercury $(II)$ chloride is treated with excess of stannous chloride,the products obtained are

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