Answer
a.
$$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + Na_3PO_4(aq)$$
b.
$$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + LiI$$
c.
$$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + Ca(NO_3)_2$$
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. $H_3PO_4$ provides 3 $H^+$, so we have to put a coefficient of 3 in front of the $NaOH$, since it only provides 1 $OH^-$.
$$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + salt$$
b. $HI$ provides 1 $H^+$, and $LiOH$ provides 1 $OH^-$. The ions are already balanced.
$$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + salt$$
c. $Ca(OH)_2$ provides 2 $OH^-$, so we have to put a coefficient of 2 in front of the $HNO_3$, since it only provides 1 $H^+$
$$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + salt$$
3. Balance the $H_2O$ with the $H^+$ and $OH^-$.
a. The reaction uses 3 $H^+$ and 3 $OH^-$. Thus, there should be 3 $H_2O$ in the products.
$$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + salt$$
b. The reaction uses 1 $H^+$ and 1 $OH^-$. Thus, there should be 1 $H_2O$ in the products.
$$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + salt$$
c. The reaction uses 2 $H^+$ and 2 $OH^-$. Thus, there should be 2 $H_2O$ in the products.
$$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + salt$$
4. Use the remaining ions to produce a salt.
a. Remaining ions: $1P{O_4}^{3-}$ and $3Na^+$:
$$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + Na_3PO_4(aq)$$
b. Remaining ions: $1I^-$ and $1Li^{+}$:
$$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + LiI$$
c. Remaining ions: $2N{O_3}^{-}$ and $Ca^{2+}$:
$$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + Ca(NO_3)_2$$
