Answer:
The volume that carbon dioxide release is 2.83L
Step-by-step explanation:
The reaction of acetic acid (CH₃COOH) with sodium carbonate (Na₂CO₃) is:
2 CH₃COOH + Na₂CO₃ →Na₂(CH₃COO)₂ + CO₂ + H₂O
Moles of acetic acid and sodium carbonate (Molar mass: 105.99g/mol) in the reaction are:
Acetic acid: 0.300L ₓ (0.83mol / L) = 0.249 moles.
Sodium carbonate: 12g ₓ (1mol / 105.99g) = 0.113 moles.
Based on the chemical equation, 2 moles of acetic acid reacts per moles of sodium carbonate. For a complete reaction of sodium carbonate you need:
0.113 moles Na₂CO₃ ₓ (2 moles CH₃COOH / 1 mole Na₂CO₃) = 0.226 moles of CH₃COOH
As you have 0.249 moles, Na₂CO₃ is limitng reactant.
As 1 mole of sodium carbonate produce 1 mole of CO₂, from 0.113 moles of Na₂CO₃ you obtain 0.113 moles of CO₂
Using PV = nRT, it is possible to find the volume that a gas occupies, thus:
V = nRT / P
n = 0.113 moles
R = 8.314 kPa×L/mol×K
T = 21°C + 273.15 = 294.15K
P = 100.3kPa - 2.49kPa = 97.81kPa
The vapor pressure is subtracted because is the pressure that water exerted.
Replacing:
V = 0.113mol×8.314 kPa×L/mol×K×294.15K / 97.81kPa
V = 2.83L
The volume that carbon dioxide release is 2.83L