Question

Liquid hexane CH 3 CH 2 4 CH 3 will react with gaseous oxygen O 2 to produce gaseous carbon dioxide CO 2 and gaseous water H 2 O . Suppose 14.6 g of hexane is mixed with 22. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

This reaction is combustion reaction. Balanced Combustion reaction is:

2C₆H₁₄ + 19O₂ (g) 12CO₂ (g) + 14H₂O(g)

Mass of Hexane = 14.6 g

Molar mass of hexane = 86.18 g/mol

Moles of hexane = mass/ molar mass = 14.6 g / 86.18 g/mol = 0.169 mol

Mass of O₂ = 22 g

Molar mass of O₂ = 32 g/mol

Moles of O₂ = 22 g / (32g/mol) = 0.6875 mol

From reaction

2 mol of Hexane reacts with 19 mol of O₂

So, 0.169 mol of Hexane reacts with 19*0.169/2 = 1.61 mol of O₂

Thus O₂ is a limiting reagent

From reaction

19 mol of O₂ produces 12 mol of CO₂

0.6875 mol of O₂ produces 12*0.6875/19 = 0.434 mol of CO₂

Molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ produced = moles * molar mass = 0.434 mol * 44.01 g/mol = 19.11 g = 19 g (sig. fig.)