A volume of 3.0 L of butane is burned in excess oxygen.Balance the chemical equation below and determine how many total liters of gases are produced.Assume that both the reactant and product temperature is 500 K and the pressure of the system remains constant at 1.0 atm.
C₄H₁₀(g) + O₂(g) → CO₂(g) + H₂O(g)

Question

Quizplus · Accepted Answer

The balanced chemical equation for the combustion of butane (C4H10) is 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g). For every 2 moles of butane burned, 18 moles of gas are produced (8 moles of CO2 and 10 moles of H2O). Since the volume of gases at the same temperature and pressure is directly proportional to the number of moles (according to Avogadro's law), burning 3.0 L of butane (1.5 times the stoichiometric amount) produces 1.5 times the gas volume, resulting in 27 L of gases (1.5 * 18 L = 27 L).

A Volume of 3