Answer to Question #174338 in Chemistry for amany

Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 90. g of octane is mixed with 145. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

CH₃CH₂₆CH₃ + 11O₂ = 3CO₂ + 16H₂O

M (CH₃CH₂₆CH₃) = 68.2862 g/mol

M (O₂) = 32 g/mol

M (H2O) = 18 g/mol

n = m / M

n (CH₃CH₂₆CH₃) = 90 / 68.2862 = 1.32 mol

n (O₂) = 145 / 32 = 4.53 mol

According to the equation, 1 mole of octane requires 11 moles of O₂. Respectively, 1.32 moles of octane require 1.32 x 11 = 14.52 moles of O₂.

Therefore, in this case, O₂ is the limiting reactant. With all the available O₂ being used, the amount of water to be produced is:

n (H₂O) =n (O₂) / 11 x 16 = 4.53 / 11 x 16 = 6.59 mol

m (H₂O) = n x M = 6.59 x 18 = 118.6 g