Question

Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and...

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.

Homework Answers

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

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