# Chapter 11 - Acids and Bases - 11.7 Reactions of Acids and Bases - Questions and Problems - Page 425: 11.58

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. $$3H_2SO_4(aq) + 2Al(OH)_3(aq) \longrightarrow 6 H_2O(l) + Al_2(SO_4)_3(aq)$$

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