Question

Gaseous ethane

CH3CH3

will react with gaseous oxygen

O2

to produce gaseous carbon dioxide

CO2

and gaseous water

H2O

. Suppose 1.80 g of ethane is mixed with 10. 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

2 CH3CH3 + 7 O2 -----> 4 CO2 + 6 H2O

number of moles of ethane = 1.80g / 30.07 g.mol^-1 = 0.0599 mole

number of moles of O2 = 10.0g / 32.0 g.mol^-1 = 0.3125 mole

from the balanced equation we can say that

2 mole of ethane requires 7 mole of O2 so

0.0599 mole of ethane will require

= 0.0599 mole of ethane *(7 mole of O2 / 2 mole of ethane)

= 0.210 mole of O2

But we have 0.3125 mole of O2

so oxygen is in excess therefore, ethane is limiting reactant

2 mole of ethane produces 6 mole of H2O

so 0.0599 mole of ethane will produce 0.1797 mole of H2O

1 mole of H2O = 18.015 g

0.1797 mole of H2O = 3.24 g

Therefore, the mass of H2O produced will be 3.24 g