Answer
a. $ZnCO_3(s) + HBr(aq) \longrightarrow CO_2(g) + H_2O(l) + ZnBr_2(aq)$
b. $Zn(s) + 2HCl(aq) \longrightarrow H_2(g) + ZnCl_2(aq)$
c. $HCl(aq) + NaHCO_3(s) \longrightarrow CO_2(g) + H_2O(l) + NaCl(aq)$
d. $H_2SO_4(aq) + Mg(OH)_2(s) \longrightarrow MgSO_4(aq) + 2H_2O(l)$
Work Step by Step
a. This is a reaction between an acid and a carbonate. First, the Carbonate ion will react with 2 protons from the acid, producing $H_2CO_3$, which will break down into $H_2O(l)$ and $CO_2(g)$. The remaining ions: $Zn^{2+}$ and $Br^-$ will produce a salt: $ZnBr_2$.
$ZnCO_3(s) + HBr(aq) \longrightarrow CO_2(g) + H_2O(l) + ZnBr_2(aq)$
b. This is a reaction between an acid and a metal. Therefore, we know that it will produce $H_2(g)$ and a salt made with the metal ion and the conjugate base of the acid.
- Metal ion: $Zn^{2+}$
- Conjugate base: $Cl^-$
Resulting salt: $ZnCl_2$
$Zn(s) + HCl(aq) \longrightarrow H_2(g) + ZnCl_2(aq)$
- Balance the equation:
$Zn(s) + 2HCl(aq) \longrightarrow H_2(g) + ZnCl_2(aq)$
c. This is a reaction between an acid and a bicarbonate. Thus, we know that it will produce $CO_2(g)$, $H_2O(l)$, using the bicarbonate and a proton from the acid, and a salt with the remaining ions.
$HCl(aq) + NaHCO_3(s) \longrightarrow CO_2(g) + H_2O(l) + NaCl(aq)$
d. This is a reaction between an acid and a hydroxide. Therefore, we know that it will produce $H_2O(l)$, with the protons from the acid, and the hydroxide ions. The remaining ions will produce a salt.
Remaining ions: $S{O_4}^{2-}$ and $Mg^{2+}$
$H_2SO_4(aq) + Mg(OH)_2(s) \longrightarrow MgSO_4(aq) + H_2O(l)$
- Balance the equation:
$H_2SO_4(aq) + Mg(OH)_2(s) \longrightarrow MgSO_4(aq) + 2H_2O(l)$
