Question

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 17. g of octane is mixed with 93.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: 52.8 g of
`CO_2`

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}=(17.0g)/(114g/mol)=0.150moles`

b) moles of oxygen

`\text{Number of moles}=(93.0g)/(32g/mol)=2.91moles`

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

According to stoichiometry :

2 moles of
`C_8H_(18)` require 25 moles of
`O_2`

Thus 0.150 moles of
`C_8H_(18)` require=
`(25)/(2)* 0.150=1.875moles` of
`O_2`

Thus
`C_8H_(18)` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.

As 2 moles of
`C_8H_(18)` give = 16 moles of
`CO_2`

Thus 0.150 moles of
`C_8H_(18)` give =
`(16)/(2)* 0.150=1.2moles` of
`CO_2`

Mass of
`CO_2=moles* {\text {Molar mass}}=1.2moles* 44g/mol=52.8g`

Thus 52.8 g of
`CO_2` will be produced from the given masses of both reactants.