Question

Liquid hexane

CH3CH24CH3

will react with gaseous oxygen

O2

to produce gaseous carbon dioxide

CO2

and gaseous water

H2O

. Suppose 6.0 g of hexane is mixed with 32.3 g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Balanced equation is

2CH₃(CH₂)₄CH₃ + 19O2 --------> 12CO2 + 14H2O

Number of moles of hexane = 6.0g / 86.18 g/mol = 0.0696 mole

Number of moles of O2 = 32.3 g / 32.0 g/mol = 1.009 mole

from the balanced equation we can say that

2 mole of hexane requires 19 mole of O2 so

0.0696 mole of hexane will require

= 0.0696 mole of hexane *(19 mole of O2 / 2 mole of hexane)

= 0.6612 mole of O2

but we have 1.009 mole of O2 which is in excess so O2 is excess reactant

and hexane is the limiting reactant so no amount of hexane will be left over in this reaction.