Gaseous ethane (CH, CH) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 24 g of

ethane is misced with 16.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2

significant digits.

Mass = 8.1 g

Explanation:

Given dа ta:

Mass of ethane = 24 g

Mass of oxygen = 16.9 g

Mass of water = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Moles of ethane:

Number of moles = mass/molar mass

Number of moles = 24 g / 30.07 g/mol

Number of moles = 0.8 mol

Moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 16.9 g / 32 g/mol

Number of moles = 0.53 mol

Now we will compare the moles of water with oxygen and ethane.

C₂H₆ : H₂O

2 : 6

0.8 : 6/2*0.8 = 2.4 mol

O₂ : H₂O

7 : 6

0.53 : 6/7*0.53 = 0.45 mol

It means oxygen is limiting reactant because moles of water produced by given amount of oxygen are less.

Mass of water:

Mass = number of moles * molar mass

Mass = 0.45 mol * 18 g/mol

Mass = 8.1 g