Write the balanced chemical equation for the reaction of sodium bicarbonate with aqueous acetic acid to produce carbon dioxide, water, and sodium acetate. Follow the example set by the sodium bicarbonate.
You didn't post the example with NaHCO3.
NaHCO3 + CH3COOH ==> CH3COONa + H2O + CO2.
NaHCO3 + CH3COOH ==> CH3COONa + H2O + CO2
To write a balanced chemical equation, we need to make sure that the number of atoms on both sides of the equation is equal. Let's break down the reaction step by step:
Step 1: Identify the reactants and products.
The reactants are sodium bicarbonate (NaHCO3) and acetic acid (CH3COOH). The products are carbon dioxide (CO2), water (H2O), and sodium acetate (CH3COONa).
Step 2: Write the unbalanced equation.
The unbalanced equation will be:
NaHCO3 + CH3COOH → CO2 + H2O + CH3COONa
Step 3: Balance the equation.
To balance the equation, we need to adjust the coefficients (numbers in front of each compound) until the number of atoms on both sides is the same.
NaHCO3 + CH3COOH → CO2 + H2O + CH3COONa
First, let's balance the carbon (C) atoms. We have one carbon atom on the left side (from CH3COOH) and one carbon atom on the right side (from CO2 and CH3COONa). So far, we have:
NaHCO3 + CH3COOH → CO2 + H2O + CH3COONa
Next, let's balance the hydrogen (H) atoms. We have three hydrogen atoms on the left side (from CH3COOH and NaHCO3) and two hydrogen atoms on the right side (from H2O). To balance the hydrogen atoms, we can place a coefficient of 2 in front of H2O:
NaHCO3 + CH3COOH → CO2 + 2H2O + CH3COONa
Finally, let's balance the sodium (Na) atoms. We have one sodium atom on the left side (from NaHCO3) and one sodium atom on the right side (from CH3COONa). The equation is now balanced:
NaHCO3 + CH3COOH → CO2 + 2H2O + CH3COONa
This is the balanced chemical equation for the reaction of sodium bicarbonate with aqueous acetic acid to produce carbon dioxide, water, and sodium acetate.