Question

Problem page liquid octane ch3ch26ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 6.9 g of octane is mixed with 10.1 g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to 3 significant digits.

Answer 1

Answer:- 8.89 g of carbon dioxide can be produced.

Solution:- The balanced equation for the combustion of octane in presence of oxygen to give carbon dioxide and water is:

`2C_8H_1_8+25O_2\rightarrow 16CO_2+18H_2O`

From given info, we have 6.9 g octane and 10.1 g oxygen and asked to calculate the maximum mass of carbon dioxide formed.

It is a stoichiometry problem. Let's first figure out the limiting reactant. For this we could either calculate grams of octane required to react completely with given grams of oxygen or vice versa.

From balanced equation, there is 25:2 mol ratio between oxygen and octane.

Let's say we calculate grams of oxygen for given grams of octane as:

`6.9g Octane((1mol)/(114g))((25mol Oxygen)/(2mol Octane))((32g Oxygen)/(1mol Oxygen))`

= 24.1 g oxygen

From calculations, 24.1 g of oxygen are required to react completely with 6.9 g of octane but in actual only 10.1 g of oxygen are available. So, oxygen is limiting reactant and so the product yield depends on it and calculated as:

`10.1g Oxygen((1mol)/(32g))((16mol Carbon dioxide)/(25mol Oxygen))((44g Carbon dioxide )/(1mol Carbondioxide))`

= 8.89 g carbon dioxide

So, from given masses of octane and oxygen, maximum 8.89 g of carbon dioxide can be produced.