Answer
a.
$$H_2SO_4(aq) + 2NaOH(aq) \longrightarrow 2H_2O(l) + Na_2SO_4(aq)$$
b.
$$3HCl(aq) + Fe(OH)_3 \longrightarrow 3H_2O(l) + FeCl_3(aq)$$
c.
$$H_2CO_3(aq) + Mg(OH)_2(aq) \longrightarrow 2H_2O(l) + MgCO_3$$
Work Step by Step
1. All neutralization reactions follow this pattern:
$$Acid + Base \longrightarrow H_2O(l) + salt$$
2. Balance the $H^+$ in the acid with the $OH^-$ in the base.
a. $H_2SO_4$ provides 2 $H^+$, so we have to put a coefficient of 2 in front of the $NaOH$, since it only provides 1 $OH^-$.
$$H_2SO_4(aq) + 2NaOH(aq) \longrightarrow H_2O(l) + salt$$
b. $Fe(OH)_3$ provides 3 $OH^-$, so we have to put a coefficient of 3 in front of the $HCl$, since it only provides 1 $H^+$
$$3HCl(aq) + Fe(OH)_3 \longrightarrow H_2O(l) + salt$$
c. $H_2CO_3$ provides 2 $H^+$, and $Mg(OH)_2$ provides 2 $OH^-$. The ions are already balanced.
$$H_2CO_3(aq) + Mg(OH)_2(aq) \longrightarrow H_2O(l) + salt$$
3. Balance the $H_2O$ with the $H^+$ and $OH^-$.
a. The reaction uses 2 $H^+$ and 2 $OH^-$. Thus, there should be 2 $H_2O$ in the products.
$$H_2SO_4(aq) + 2NaOH(aq) \longrightarrow 2H_2O(l) + salt$$
b. The reaction uses 3 $H^+$ and 3 $OH^-$. Thus, there should be 6 $H_2O$ in the products.
$$3HCl(aq) + Fe(OH)_3 \longrightarrow 3H_2O(l) + salt$$
c. The reaction uses 2 $H^+$ and 2 $OH^-$. Thus, there should be 2 $H_2O$ in the products.
$$H_2CO_3(aq) + Mg(OH)_2(aq) \longrightarrow 2H_2O(l) + salt$$
4. Use the remaining ions to produce a salt.
a. Remaining ions: $1S{O_4}$ and $2Na^+$:
$$H_2SO_4(aq) + 2NaOH(aq) \longrightarrow 2H_2O(l) + Na_2SO_4(aq)$$
b. Remaining ions: $3Cl^+$ and $1Fe^{3+}$:
$$3HCl(aq) + Fe(OH)_3 \longrightarrow 3H_2O(l) + FeCl_3(aq)$$
c. Remaining ions: $C{O_3}^{2-}$ and $Mg^{2+}$:
$$H_2CO_3(aq) + Mg(OH)_2(aq) \longrightarrow 2H_2O(l) + MgCO_3$$
