Answer
a. No reaction.
b. No reaction.
c. $CrBr_2(aq) + Na_2CO_3(aq) \longrightarrow CrCO_3(s) + 2 NaBr(aq)$
d. $3NaOH(aq) + FeCl_3(aq) \longrightarrow 3 NaCl(aq) + Fe(OH)_3(s)$
Work Step by Step
1. Determine the resulting ionic compounds by switching the anions (or cations):
a. $Li_2S$ and $BaI_2$
b. $K_2S$ and $CaCl_2$
c. $CrCO_3$ and $NaBr$
d. $NaCl$ and $Fe(OH)_3$
2. Determine if they are soluble or insoluble.
a. Both are soluble.
b. Both are soluble.
c. $CrCO_3$ is insoluble.
d. $Fe(OH)_3$ is insoluble.
3. 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. No reaction.
b. No reaction.
c. $CrBr_2(aq) + Na_2CO_3(aq) \longrightarrow CrCO_3(s) + 2 NaBr(aq)$
d. $3NaOH(aq) + FeCl_3(aq) \longrightarrow 3 NaCl(aq) + Fe(OH)_3(s)$
