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What is the percent yield when 7.80 grams of benzene burns in oxygen gas to form 3 grams of CO2 and water vapor?
Wiki User
∙ 12y ago
C6H6 + 15 O2 = 12 CO2 + 6 H2O
moles of C6H6 = 7.8 / 78 = .1 moles
the O2 is in excess so benzene is the limiting agent and forms .1 x 6 moles of CO2
.6 moles of CO2 = .6 x 44 = 26.7 g at 100 % yield
but we have 3 g or 3 / 26.7 % = 11.23 %
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∙ 12y ago
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