Question

Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 62. g of octane is mixed with 100. 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 = 40.5 g

Step-by-step explanation:

Given data:

Mass of liquid octane = 62.0 g

Mass of oxygen = 100 g

Mass of water produced = ?

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Number of moles of octane:

Number of moles = mass/molar mass

Number of moles = 62.0 g/114.23 g/mol

Number of moles = 0.54 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 100 g/ 32 g/mol

Number of moles = 3.12 mol

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

C₈H₁₈ : H₂O

2 : 18

0.54 : 18/2×0.54 = 4.86 mol

O₂ : H₂O

25 : 18

3.12 : 18/25×3.12 = 2.25 mol

Number of moles of water produced by oxygen are less so it will act as limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 2.25 mol × 18 g/mol

Mass = 40.5 g