Butane (C4H10) is used as a fuel where natural gas is not available. How many grams of butane will fill a 3.50-liter container a

Question

3 years ago

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Butane (C4H10) is used as a fuel where natural gas is not available. How many grams of butane will fill a 3.50-liter container a

t 35.6 °C and 758 torr? I've put the values through n = PV/RT to get n = 0.0138 and then converted to grams to get 0.77423g but it's not the right answer on my practice midterm. Any help is gratefully appreciated!

Chemistry

1 answer:

erik [133] · Answer 1 · 2022-06-08T05:10:14+03:00

erik [133]3 years ago

6 0

<span>We use the formula PV = nRT. P = 758 torr = 0.997 atm. V = 3.50 L. T = 35.6 C = 308.15 K. R = 0.0821. Rearranging the equation gives up n = PV/Rt and we get .0138 moles of butane. Mass of 0.0138 moles of butane = .0138 x 58.12 = 8.02g.</span>