Calculate the minimum mass

Question

Liquid hexane (CH3(CH2)4CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide(CO2) and gaseous water (H2O).Suppose 7.76 g of hexane is mixed with 45. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to significant digits.

Answer 1

First, you must write the correctly balanced equation for the reaction of hexane with oxygen:

2C₆H₁₄(l) + 13O₂(g) ==> 12CO₂(g) + 14H₂O(g) ... balanced equation

Next we need to find which reactant is limiting. An easy way is to divide mols of each reactant by the corresponding coefficient in the balanced equation:

-- For C₆H₁₄: 7.76 g x 1 mol/86.18 g = 0.0900 mols C₆H₁₄ (÷2->0.045)

-- For O₂: 45. g x 1 mol/32 g = 1.406 mols O₂ (÷13->0.108)

Since 0.045 is less than 0.108, C₆H₁₄ is limiting and will run out first before all the oxygen is used up.

Theoretically, there should be NO hexane left over. Perhaps the question meant to ask for the minimum mass of OXYGEN that could be left over.

moles O₂ left over = initial mols O₂ - mols O₂ used

mols O₂ used = 0.0900 mols C₆H₁₄ x 13 mols O₂ / 2 mols C₆H₁₄ = 0.585 mols O₂ used

mols O₂ left over = 1.406 mols O₂ - 0.585 mols O₂ = 0.82 mols O₂ left over (2 sig. figs.)