Answer
a. $$2HNO_3(aq) + Ba(OH)_2(aq) \longrightarrow 2H_2O(l) + Ba(NO_3)_2(aq)$$
b.
$$3H_2SO_4(aq) + 2Al(OH)_3(aq) \longrightarrow 6 H_2O(l) + Al_2(SO_4)_3(aq)$$
Work Step by Step
1. Balance the $H^+$ in the acid with the $OH^-$ in the base.
a. $Ba(OH)_2$ provides 2 $OH^-$, so we have to put a coefficient of 2 in front of the $HNO_3$, since it only provides 1 $H^+$
$$2HNO_3(aq) + Ba(OH)_2(aq) \longrightarrow H_2O(l) + Ba(NO_3)_2(aq)$$
b. $H_2SO_4$ provides 2 $H^+$, and $Al(OH)_3$ provides 3 $OH^-$, so we have to put a coefficient of 3 in front of the $H_2SO_4$, and 2 in front of $Al(OH)_3$.
$$3H_2SO_4(aq) + 2Al(OH)_3(aq) \longrightarrow H_2O(l) + Al_2(SO_4)_3(aq)$$
2. 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.
$$2HNO_3(aq) + Ba(OH)_2(aq) \longrightarrow 2H_2O(l) + Ba(NO_3)_2(aq)$$
b. The reaction uses 6 $H^+$ and 6 $OH^-$. Thus, there should be 6 $H_2O$ in the products.
$$3H_2SO_4(aq) + 2Al(OH)_3(aq) \longrightarrow 6 H_2O(l) + Al_2(SO_4)_3(aq)$$
