To answer this question, we first need to determine the balanced chemical equation for the combustion reaction between propane (C3H8) and oxygen (O2). The balanced equation is:
C3H8 + 5O2 -> 3CO2 + 4H2O
According to the balanced equation, one molecule of propane reacts with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water.
To calculate the volume of carbon dioxide gas formed, we need to use stoichiometry and the given volume of propane and oxygen. Based on the balanced equation, we can see that the ratio between propane and carbon dioxide is 1:3.
Step 1: Convert the given volumes to moles using the ideal gas law equation, PV = nRT. Assuming standard temperature and pressure (STP), where T = 273 K and P = 1 atm, the equation simplifies to PV = n.
For propane (C3H8):
n(C3H8) = (20 L)(1 atm) / (22.4 L/mol) --> 0.8928 mol
For oxygen (O2):
n(O2) = (55.7 L)(1 atm) / (22.4 L/mol) --> 2.4879 mol
Step 2: Determine the limiting reactant. The balanced equation shows that 1 mole of propane reacts with 5 moles of oxygen. Therefore, for 0.8928 moles of propane, we need 4.464 moles of oxygen. Since we have 2.4879 moles of oxygen, oxygen is in excess.
Step 3: Calculate the moles of carbon dioxide formed using the ratio from the balanced equation. According to the equation, for every 1 mole of propane reacted, 3 moles of carbon dioxide are produced.
n(CO2) = (0.8928 mol C3H8) * (3 mol CO2 / 1 mol C3H8) --> 2.6784 mol CO2
Step 4: Convert the moles of carbon dioxide to volume using the ideal gas law equation.
V(CO2) = n(CO2) * (22.4 L/mol) --> V(CO2) = 2.6784 mol * (22.4 L/mol) --> V(CO2) = 59.9 L
Therefore, when 20 L of propane reacts with 55.7 L of oxygen, approximately 59.9 L of carbon dioxide gas is formed.