Q. vinegar + baking soda equation

Step 1. Identify the chemical species. Vinegar is an aqueous solution of acetic acid, with formula \( \mathrm{CH_3COOH} \). Baking soda is sodium bicarbonate, with formula \( \mathrm{NaHCO_3} \). The common products of their reaction are sodium acetate, water, and carbon dioxide, with formulas \( \mathrm{CH_3COONa} \), \( \mathrm{H_2O} \), and \( \mathrm{CO_2} \) respectively.

Step 2. Write the unbalanced equation using those formulas. Use an equals sign instead of an arrow.
\[ \mathrm{CH_3COOH} + \mathrm{NaHCO_3} = \mathrm{CH_3COONa} + \mathrm{H_2O} + \mathrm{CO_2} \]

Step 3. Count atoms of each element on the reactant side. For the left side:
\( \mathrm{CH_3COOH} \) contributes \( C=2 \), \( H=4 \), \( O=2 \), \( \) and \( \mathrm{NaHCO_3} \) contributes \( \mathrm{Na}=1 \), \( C=1 \), \( H=1 \), \( O=3 \).
So totals on the left are
\( C_{\text{left}} = 2 + 1 = 3 \),
\( H_{\text{left}} = 4 + 1 = 5 \),
\( O_{\text{left}} = 2 + 3 = 5 \),
\( \mathrm{Na}_{\text{left}} = 1 \).

Step 4. Count atoms on the product side. For the right side:
\( \mathrm{CH_3COONa} \) gives \( C=2 \), \( H=3 \), \( O=2 \), \( \mathrm{Na}=1 \).
\( \mathrm{H_2O} \) gives \( H=2 \), \( O=1 \).
\( \mathrm{CO_2} \) gives \( C=1 \), \( O=2 \).
So totals on the right are
\( C_{\text{right}} = 2 + 1 = 3 \),
\( H_{\text{right}} = 3 + 2 = 5 \),
\( O_{\text{right}} = 2 + 1 + 2 = 5 \),
\( \mathrm{Na}_{\text{right}} = 1 \).

Step 5. Compare the totals. Each element has the same total on both sides:
\( C: 3 = 3 \),
\( H: 5 = 5 \),
\( O: 5 = 5 \),
\( \mathrm{Na}: 1 = 1 \).
Therefore the equation as written is already balanced with all stoichiometric coefficients equal to 1.

Step 6. Present the balanced chemical equation.
\[ \mathrm{CH_3COOH} + \mathrm{NaHCO_3} = \mathrm{CH_3COONa} + \mathrm{H_2O} + \mathrm{CO_2} \]

Step 7. Brief note on observation. Carbon dioxide is released as bubbles, water remains in solution, and sodium acetate stays dissolved if the reaction is in water.