Answer : No octane could be left over by the chemical reaction.
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of octane
[tex]\text{Number of moles}=\frac{10g}{114g/mol}=0.088moles[/tex]
b) moles of oxygen
[tex]\text{Number of moles}=\frac{61.9g}{32g/mol}=1.9moles[/tex]
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
According to stoichiometry :
2 moles of octane require 25 moles of oxygen
Thus 0.088 moles of octane require =[tex]\frac{25}{2}\times 0.088=1.1moles[/tex] Â 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.
