Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.3 g of octane is mixed with 12.4 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Answer: Octane will be used completely.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of octane}=(2.3g)/(114g/mol)=0.0202moles`

`\text{Moles of oxygen}=(12.4g)/(32g/mol)=0.388moles`

The balanced chemical reaction will be

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

According to stoichiometry :

2 moles of octane require = 25 moles of
`O_2`

Thus 0.0202 moles of octane will require=
`(25)/(2)* 0.0202=0.2525moles` of
`O_2`

Thus octane is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.

Thus octane will be used completely.