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

Mass of water will be 3.24 g.

Step-by-step 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