1. **State the problem:** We need to find the mass of water formed when 145.0 mL of 1.5 M acetic acid reacts with 195.0 mL of 1.25 M sodium bicarbonate. This is a limiting reactant problem. 2. **Write the balanced chemical equation:** $$\mathrm{HC_2H_3O_2 (aq) + NaHCO_3 (aq) \rightarrow NaC_2H_3O_2 (aq) + H_2O (l) + CO_2 (g)}$$ 3. **Calculate moles of each reactant:** - Moles of acetic acid: $$n_{\mathrm{acid}} = M \times V = 1.5 \times \frac{145.0}{1000} = 0.2175$$ moles - Moles of sodium bicarbonate: $$n_{\mathrm{bicarb}} = 1.25 \times \frac{195.0}{1000} = 0.24375$$ moles 4. **Determine the limiting reactant:** The reaction ratio is 1:1, so the limiting reactant is the one with fewer moles. $$0.2175 < 0.24375 \Rightarrow \text{acetic acid is limiting}$$ 5. **Calculate moles of water formed:** From the balanced equation, 1 mole of acetic acid produces 1 mole of water. $$n_{\mathrm{H_2O}} = 0.2175$$ moles 6. **Calculate mass of water:** Molar mass of water $$= 18.015\,g/mol$$ $$m = n \times M = 0.2175 \times 18.015 = 3.916$$ g 7. **Final answer:** The mass of water formed is approximately **3.9 g**.