Limiting reactants

Question

Liquid hexane
CH3CH24CH3
will react with gaseous oxygen
O2
to produce gaseous carbon dioxide
CO2
and gaseous water
H2O
. Suppose 2.6 g of hexane is mixed with 5.29 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to
3
significant digits.

Answer 1

hexane is C6H14 or put another way, as in the question CH3(CH2)4CH3. Actually, it is incorrect as depicted in the question.

 

Nonetheless...

2C6H14 + 19O2 ===> 12CO2 + 14H2O ... balanced equation

 

moled hexane present = 2.6 g x 1 mole/130 g = 0.02 moles

moles O2 present = 5.29 g x 1 mole/32 g = 0.165 moles

Which reactant is limiting? Hexane = 0.02/2 = 0.01; O2 = 0.165/19 = 0.0087

Thus O2 is limiting...

moles of H2O that can be produced =0.165 moles O2 x 14 H2O/19 CO2 = 0.122 moles H2O

Mass H2O = 0.122 moles x 18 gm/mole = 2.20 g (to 3 sig. figs.)