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A bottle is labelled '112% H2SO4' (oleum). When excess water is added to it and the resulting solution is reacted with 5.3 g of Na2CO3, what volume of CO2 gas is produced at 1 atm and 300 K after the reaction goes to completion?

1. 2.46 L
2. 24.6 L
3. 1.23 L
4. 12.3 L

Correct answer: 1.23 L

Solution

112% oleum means per 100 g sample there are 12 g of free SO3 (which equals 0.15 mol SO3 -> 0.15 mol H2SO4 extra) and 100 g H2SO4 base. Na2CO3 (5.3 g = 0.05 mol) reacts 1:1 with H2SO4 to release 0.05 mol CO2; at 1 atm, 300 K, V = (0.05)(0.0821)(300) = 1.23 L.

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