Question

Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 2.80g of carbon dioxide is produced from the reaction of 3.4g of octane and 5.9g of oxygen gas, calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

Answer 1

Answer:

54%

Explanation:

The balanced equation is:

$2 CH_{3}(CH_{2})_{6}CH_{3} + 25 O_{2} = 16 CO_{2} + 18 H_{2}O$

The first step is to determine the limiting reactant. For this, we calculate the moles of each given component and divide the result for the stoichiometric coefficient.

3.4 g octane / 114.23 g/mol = 0.030 mol octane

0.030 mol octane/2=0.015

5.9 g O2 / 32 g/mol = 0.18 mol O2

0.18 mol O2/25= 0.0074 mol

The lower number, in this case oxygen, is the limiting reactant. The value corresponds to the theoretical yield of the reaction.

Similarly, the real yield is calculated from the product.

2.80 g CO2/ 44.01 g/mol = 0.0636 mol CO2

0.0636 mol CO2/16 = 0.00398 mol

The percent yield is the ratio of the 2 multiplied by a hundred, then

Percent yield= 0.0398/0.0074 *100 = 54%