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.

Mass = 40.5 g

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

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.