Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.28 g of octane is mixed with 15. 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

Answer:

Mass of water will be 3.24 g.

Explanation:

Given data:

mass of octane = 2.28 g

mass of oxygen = 15 g

mass of water = ?

Solution:

First of all we will write the balance chemical equation.

2C₈H₁₈ + 25O₂     →     16CO₂ + 18H₂O

Now we will calculate the moles of octane and oxygen.

number of moles of octane = mass / molar mass

number of moles of octane = 2.28 g / 114.33 g/mol

number of moles of octane =  0.02 mol

number of moles of oxygen = mass / molar mass

number of moles of oxygen = 15 g/ 32 g/mol

number of moles of oxygen =0.47 mol

Now we will compare the moles of oxygen and octane with water from balance chemical equation

                     C₈H₁₈             :       H₂O

                       2                  :         18

                      0.02             :         18/2×0.02 = 0.18 mol

 

                      O₂                :          H₂O

                        25             :            18

                        0.47        :              18/25×0.47 = 0.34 mol

Number of moles of water produced by octane is less that's why octane will be the limiting reactant.

       Mass of water = number of moles × molar mass

       Mass of water =  0.18 mol × 18 g/mol = 3.24 g