Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 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.

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 > 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2