Question

Gaseous ethane (CH3CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 26. g of ethane is mixed with 180. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

  CH₃CH₃(g) + 3.5O₂(g) ----> 2CO₂(g) + 3H₂O(l)

Molar mass of ethane = 30.07 g/mol

Molar mass of oxygen = 16 g/mol

Molar mass of water = 18.02 g/mol

(For multiplication/division answer should have least number of significant figures )

Mass of ethane = 26 g (2 significant digits)

Mass of oxygen = 180 g (2 significant digits)

No. of moles of ethane given = mass/molar mass = 26g/30.07 g/mol = 0.87 moles (2 significant digits)

No. of moles of oxygen given = mass/molar mass = 180 g/18.02 g/mol = 10 moles (2 significant digits)

Molar ratio of ethane and oxygen = 1:3.5

1 mole of ethane require 3.5 moles of oxygen

0.865 moles of ethane require == 0.87 x 3.5 = 3.0 moles (2 significant digits)

Given moles of oxygen = 10 moles (2 significant digits)

This shows that ethane is limiting reactant

Molar ratio of ethane and water = 1:3

1 mole of ethane produce 3 moles of water

0.87 moles of ethane produce ==> 0.87 x 3 = 2.6 moles (2 significant digits)

Maximum mass of water that can produced = theoretical moles x molar mass = 2.6 mol x 18.02 g/mol = 47 g