16.8 g of sodium hydrogen trioxocarbonate (IV) is completely decomposed by heat. Calculate the volume of carbon(IV) oxide given off at s.t.p
[Na 23, C = 12, O = 16, H = 1, Molar volume of a gas at s.t.p = 22.4 dm3]
- 22.40 dm3
- 11.20 dm3
-
2.24 dm3
- 1.12 dm3
The correct answer is: C
Explanation
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2
Relative molecular mass of NaHCO3 = 23 +1 + 12 + 16 * 3 = 84gmol-1
2 * 84g of NaHCO3 will be liberated 22.4dm3 or CO2 at s.t.p
16.8g of NaHCO3 will liberate (16.8 * 22.4)/(2 * 84) = 2.24 dm3
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