Answer:
1.73g of CO2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
NaHCO3 + CH3COOH → CH3COONa + H2O + CO2
Next we shall determine the masses of NaHCO3 and CH3COOH that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:
Molar mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 84g/mol
Mass of NaHCO3 from the balanced equation = 1 x 84 = 84g
Molar mass of CH3COOH = 12 + (3x1) + 12 + 16 + 16 + 1 = 60g/mol
Mass of CH3COOH from the balanced equation = 1 x 60 = 60g
Molar mass of CO3 = 12 + (2x16) = 44g/mol
Mass of CO2 from the balanced equation = 1 x 44 = 44g
From the balanced equation above,
84g of NaHCO3 reacted with 60g of CH3COOH to produce 44g of CO2.
Next, we shall determine the limiting reactant of the reaction. This is illustrated below:
From the balanced equation above,
84g of NaHCO3 reacted with 60g of CH3COOH.
Therefore, 3.3g of NaHCO3 will react with = (3.3 x 60)/84 = 2.36g of CH3COOH.
From the above illustration, we can see that only 2.36g of CH3COOH out of 10.3g given reacted completely with 3.3g of NaHCO3. Therefore, NaHCO3 is the limiting reactant while CH3COOH is the excess reactant.
Finally, can determine the mass of CO2 produced during the reaction.
In this case the limiting reactant will be used because it will produce the mass yield of CO2 as all of it were used up in the reaction. The limiting reactant is NaHCO3 and the mass of CO2 produced is obtained as shown below:
From the balanced equation above,
84g of NaHCO3 reacted to produce 44g of CO2.
Therefore, 3.3g of NaHCO3 will react to produce = (3.3 x 44)/84 = 1.73g of CO2.
Therefore, 1.73g of CO2 is released during the reaction.
