Question

Consider the unbalanced equation for the neutralization of acetic acid

Answer 1

Consider the following balanced equation for the neutralization of acetic acid.

2HC2H3O2(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(C2H3O2)2(aq)

Determine how many moles of Ca(OH)2 are required to completely neutralize 8.67 mol of HC2H3O2.

Answer 2

Rewrite the skeletal reaction between acetic acid and barium hydroxide as follows:

$CH_3COOH(aq)+Ba(OH)_2(aq)$⇒$H_2O(l)+Ba(CH_3COO)_2(aq)$

Here, the number of acetate ions $(CH_3COO^-)$ in the reactants side is 1 and the number of acetate ions in the products side is 2. So to balance them, place a coefficient 2 before $CH_3COOH$.

This gives us two moles of $H^+$ and $OH^-$ in the reactants side, that is two moles of water. But there is only one water molecule in the products side, so place a coefficient 2 before $H_2O(l)$.

Thus, the "balanced" equation is as follows:

$2CH_3COOH(aq)+Ba(OH)_2(aq)$⇒$2H_2O(l)+Ba(CH_3COO)_2(aq)$