How many Liters of I2 are produced when 241.6 grams of KI are reacted with excess CuBr2?
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The percent yield of carbon dioxide will be 49.0 %.
<h3>Percent yield</h3>First, let's look at the equation of the reaction:
The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
More on percent yield can be found here: brainly.com/question/17042787
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