300mL of 0.83mol/L acetic acid reacts with 12.0g of sodium carbonate at 21 C and 100.3kPa. What volume of dry carbon dioxide is released in this reaction? Water vapour pressure at this temperature is 2.49 kPa.

The volume that carbon dioxide release is 2.83L

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

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