Answer
a. $2NaCl(aq) + Pb(C_2H_3O_2)_2(aq) \longrightarrow 2NaC_2H_3O_2(aq) + PbCl_2(s)$
b. No reaction.
c. $2CsCl(aq) + CaS(aq) \longrightarrow Cs_2S(s) + CaCl_2(aq)$
d. $Cr(NO_3)_3(aq) + Na_3PO_4(aq) \longrightarrow 3 NaNO_3(aq) + CrPO_4(s)$
Work Step by Step
1. Write the molecular formula of each compound.
a. $NaCl$ and $Pb(C_2H_3O_2)_2$
b. $K_2S$ and $SrI_2$
c. $CsCl$ and $CaS$
d. $Cr(NO_3)_3$ and $Na_3PO_4$
2. Determine the resulting ionic compounds by switching the anions (or cations):
a. $NaC_2H_3O_2$ and $PbCl_2$
b. $KI$ and $SrS$
c. $Cs_2S$ and $CaCl_2$
d. $NaNO_3$ and $CrPO_4$
3. Determine if they are soluble or insoluble.
a. $PbCl_2$ is insoluble.
b. Both are soluble.
c. $Cs_2S$ is insoluble.
d. $CrPO_4$ is insoluble.
4. If both compounds are soluble, there will be no reaction. If some are insoluble, there will occur their precipitation. Do not forget to balance the reaction.
a. $2NaCl(aq) + Pb(C_2H_3O_2)_2(aq) \longrightarrow 2NaC_2H_3O_2(aq) + PbCl_2(s)$
b. No reaction.
c. $2CsCl(aq) + CaS(aq) \longrightarrow Cs_2S(s) + CaCl_2(aq)$
d. $Cr(NO_3)_3(aq) + Na_3PO_4(aq) \longrightarrow 3 NaNO_3(aq) + CrPO_4(s)$
