Question

Solving for row 4 For column B we are assuming that the concentration of acetic acid was 0.42 moles and the volume of acetic acid was 0.050 L

Answer 1

1) Set the chemical equation.

Acetic acid: CH3COOH

Sodium bicarbonate: NaHCO3

$CH_3COOH+NaHCO_3=CH_3COONa+H_2O+CO_2$

2) Moles of CH3COOH

$\text{Molarity CH}_3COOH\text{ =}\frac{molesofCH_3COOH}{\text{volume (L)}}$$0.42M\text{ =}\frac{molesofCH_3\text{COOH}}{0.050\text{ L}}$$0.42M\cdot0.050\text{ L = }\frac{molesofCH_3COOH}{0.050\text{ L}}\cdot0.050\text{ L}$

Moles of CH3COOH = 0.42*0.050 L = 0.021 moles of CH3COOH.