Grams of butane (C4H10) formed by the liquefaction of 448 litres of the gas (measured at (STP) would be__________________?
Correct answer: D. 1160
- A. 580
- B. 640
- C. Data insufficient; can't be computed
- D. 1160
Explanation
At STP, 448 litres corresponds to 448/22.4 = 20 moles of butane. Its mass is 20 multiplied by its molar mass of 58 g/mol, giving 1160 g.
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