Respuesta :
Explanation:
a. [tex]Pb(NO_3)_2(aq) + Na_2SO_4(aq)[/tex] → ?
[tex]Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)[/tex]
[tex]Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)[/tex]
[tex]Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)[/tex]
[tex]Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)[/tex]
Removing common ions from both sides, we get the net ionic equation:
[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]
b. [tex]NiCl_2(aq) + NH_4NO_3(aq) [/tex] →
[tex]NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)[/tex]
No precipitation is occuring.
c. [tex]Fe_Cl2(aq) + Na_2S(aq)[/tex] →
[tex]FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)[/tex]
[tex]FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)[/tex]
[tex]Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)[/tex]
[tex]Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)[/tex]
Removing common ions from both sides, we get the net ionic equation:
[tex]Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)[/tex]
d.[tex] MgSO_4(aq) + BaCl_2(aq) [/tex] →
[tex]MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2[/tex]
[tex]MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)[/tex]
[tex]BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)[/tex]
[tex]Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)[/tex]
Removing common ions from both sides, we get the net ionic equation:
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]