Answer:
Balanced chemical equation:
\(HgCl_2_(_a_q_)~+~K_2S_(_a_q_)->~2KCl_(_a_q_)~+~HgS_(_s_)\)
Overall ionic equation:
\(Hg^+^2_(_a_q_)~+~Cl^-^1~_(_a_q_)~+~K^+^1_(_a_q_)~+~S^-^2_(_a_q_)->~K^+^1_(_a_q_)~+~Cl^-^1_(_a_q_)~+~HgS_(_s_)\)
Spectator ions
\(Cl^-^1~_(_a_q_)\)
\(K^+^1_(_a_q_)\)
Possible precipitates
\(HgS_(_s_)\)
Net ionic equation
\(Hg^+^2_(_a_q_)~+~S^-^2_(_a_q_)->~HgS_(_s_)\)
Explanation:
For the chemical equation we will have:
\(HgCl_2_(_a_q_)~+~K_2S_(_a_q_)->~2KCl_(_a_q_)~+~HgS_(_s_)\)
If we follow the solubility rules, we can find the state of the products KCl and HgS. For the case of KCl we will have an aqueosus state since all the salts of \(K^+\) are soluble. For HgS we will have a solid state because all the sulfide salts are non-soluble (except the ones made of \(K^+~Na^+~NH_4^+\)).
For the overall ionic equation we will have:
\(Hg^+^2_(_a_q_)~+~Cl^-^1~_(_a_q_)~+~K^+^1_(_a_q_)~+~S^-^2_(_a_q_)->~K^+^1_(_a_q_)~+~Cl^-^1_(_a_q_)~+~HgS_(_s_)\)
We have to keep in mind that the compounds that can be broken into his ions are the ones that have the aqueous state (aq).
For the spectator ions we have to search the compounds that are in both sides:
\(Cl^-^1~_(_a_q_)\)
\(K^+^1_(_a_q_)\)
For the net ionic equation we have to remove the spectator ions:
\(Hg^+^2_(_a_q_)~+~S^-^2_(_a_q_)->~HgS_(_s_)\)
