How many moles of pure Na2CO3 were there in the sample?

2017-10-31 10:44 am

A student-made sample of sodium carbonate weighing 0.169g that generates 37.6 mL of CO2 gas when treated with sulfuric acid. The atmospheric pressure in the lab is 731.8 mm Hg and the temp. is 21.3-degree Celsius.

回答 (1)

胡雪8°

2017-10-31 11:43 am

✔ 最佳答案

Consider the CO₂ formed in the reaction :
Pressure, P = 731.8/760 atm
Volume, V = 37.6 mL = 0.0376 L
Gas constant, R = 0.0821 atm L / (mol K)
Temperature, T = (273.2 + 21.3) K = 294.5 K

Gas law : PV = nRT
Then, n = PV/(RT)

No. of moles of CO₂ formed, n = (731.8/760) × 0.0376 / (0.0821 × 294.5) mol = 0.001497 mol

The balanced equation for the reaction :
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂
Mole ratio Na₂CO₃ : CO₂ = 1 : 1
No. of moles of Na₂CO₃ reacted = 0.001497 mol

Molar mass of Na₂CO₃ = (23.0×2 + 12.0 + 16.0×2) g/mol = 106.0 g/mol
Mass of Na₂CO₃ reacted = (0.001497 mol) × (106.0 g/mol) = 0.159 g

Percent of pure Na₂CO₃ in the sample = (0.159/0.169) × 100% = 94.1%

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