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liquid octane(CH3)(CH2)6CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water(H2O). If 5.18 g of carbon dioxide is produced from the reaction of 3.43 g of octane and 19.1 g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 significant figures.

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Oseni

The percent yield of carbon dioxide will be 49.0 %.

Percent yield

First, let's look at the equation of the reaction:

[tex]2C_8H_1_8 + 25O_2 -- > 16CO_2 + 18H_2O[/tex]

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: https://brainly.com/question/17042787

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