Given:
- Mass of sodium bicarbonate (NaHCO3) = 42g
- Mass of ethanoic acid (CH3COOH) = 30g
- Mass of sodium ethanoate (CH3COONa) = 41g
- Mass of water (H2O) = 9g
- Mass of carbon dioxide (CO2) = x grams (to be calculated)

Step 1: Balanced Chemical Equation
The reaction between sodium bicarbonate and ethanoic acid can be represented by the following balanced chemical equation:
2CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2

Step 2: Molar Mass Calculation
- Molar mass of sodium bicarbonate (NaHCO3) = 22.99 + 1.01 + 12.01 + (3 * 16.00) = 84.01 g/mol
- Molar mass of ethanoic acid (CH3COOH) = (2 * 12.01) + (4 * 1.01) + (2 * 16.00) = 60.05 g/mol
- Molar mass of sodium ethanoate (CH3COONa) = (2 * 12.01) + (3 * 1.01) + 22.99 = 82.03 g/mol
- Molar mass of water (H2O) = (2 * 1.01) + 16.00 = 18.02 g/mol
- Molar mass of carbon dioxide (CO2) = (12.01 + 2 * 16.00) = 44.01 g/mol

Step 3: Stoichiometric Calculation
Using the balanced chemical equation, we can determine the number of moles of reactants and products involved.

Moles of sodium bicarbonate = Mass / Molar mass = 42g / 84.01 g/mol = 0.50 mol
Moles of ethanoic acid = Mass / Molar mass = 30g / 60.05 g/mol = 0.50 mol

According to the balanced equation, 2 moles of ethanoic acid react with 1 mole of sodium bicarbonate to produce 1 mole of carbon dioxide. Therefore, 0.50 moles of ethanoic acid will react with 0.25 moles of sodium bicarbonate to produce 0.25 moles of carbon dioxide.

Moles of carbon dioxide = 0.25 mol

Step 4: Calculation of Mass of Carbon Dioxide
Mass of carbon dioxide = Moles x Molar mass = 0.25 mol x 44.01 g/mol = 11.0025 g

Therefore, the value of x (mass of carbon dioxide) is approximately 11 grams.

Answer:
The value of x, which represents the mass of carbon dioxide formed, is approximately 11 grams.