2 NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
Explanation:
The most probably chemical reaction is:
2 NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
Sodium Hydroxide (solid) + Carbon Dioxide (gas) → Sodium Carbonate (solid) + Water (gas)
In this reaction the water you obtain have to be in liquid form because the reaction takes place at room temperature. However because the student left the beaker for 24 hours, the water have a plenty of time to evaporate, so in the end only sodium carbonate (Na₂CO₃) remained in the beaker.
number of moles = mass / molar weight
number of moles of NaOH = 0.4 / 40 = 0.01 moles
Taking in account the chemical reaction, we devise the following reasoning:
if 2 moles of NaOH produce 1 mole of Na₂CO₃
then 0.01 moles of NaOH produce X mole of Na₂CO₃
X = (0.01 × 1) / 2 = 0.005 moles of Na₂CO₃
mass = number of moles × molar weight
mass of Na₂CO₃ = 0.005 × 106 = 0.53 g
mass of Na₂CO₃ = mass of beaker and its contents after 24h - mass of empty beaker
mass of Na₂CO₃ = 113.09 - 112.58 = 0.51 g which is close the theoretical value of 0.53 g.
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