Question

Problem PageQuestion Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10. g of octane is mixed with 61.9 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer : No octane could be left over by the chemical reaction.

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) moles of octane

`\text{Number of moles}=(10g)/(114g/mol)=0.088moles`

b) moles of oxygen

`\text{Number of moles}=(61.9g)/(32g/mol)=1.9moles`

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

According to stoichiometry :

2 moles of octane require 25 moles of oxygen

Thus 0.088 moles of octane require =
`(25)/(2)* 0.088=1.1moles` of ethane

Thus octane is the limiting reagent as it limits the formation of product. Oxygen is the excess reagent as it is left in the reaction.

Thus octane will be completely used in the reaction.

Thus no octane could be left over by the chemical reaction.