When 14.2 grams of butane are burned, how many grams of oxygen are needed?
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Expert's answer
2013-03-01T04:19:11-0500
The chemical equation for theburning reaction is 2C4H10 + 13O2 = 8CO2 + 10H2O n(C4H10) = m(C4H10) / MW(C4H10) = m(C4H10) / [4 x MW(C) + 10 xMW(H)] = 14.2 / (4 x 12 + 10 x 1) = 0.245 mol According to chemical equation n(O2) = n(C4H10) : 2 x 13 = 0.245 : 2 x 13 = 1.59 mol m(O2) = n(O2) x MW(O2) = 1.59 x 32 = 50.88 g Answer: m(O2) = 50.88 g
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