Propane reacts with oxygen in a combustion reaction. What volume of carbon dioxide gas is formed when 20 L of propane are reacted with 55.7 L of oxygen?
C3H8 + 5O2 ==> 3CO2 + 4H2O
When gases are involved, we can omit the conversion to mols stage and use only volumes.
Convert 20L C3H8 to L CO2. 20 x (3 mol CO2/1 mol C3H8) = 60 L C3H8.
Convert 55.7 L O2 to CO2. 20 x (3 mols CO2/5 mols O2) = 33.4 L CO2.
One of these answers must be wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, 33.4 L CO2 will be produced. There is some unreacted propane left over.
To determine the volume of carbon dioxide gas formed when propane reacts with oxygen, we need to use the balanced equation for the combustion of propane:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
From the balanced equation, we can see that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide. Therefore, we need to find the number of moles of propane and oxygen and use the mole ratio to find the moles of carbon dioxide gas formed.
First, let's calculate the number of moles of propane:
Molar mass of propane (C₃H₈) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.1 g/mol
20 L of propane at standard temperature and pressure (STP) is equal to 20 L * (1 mol/22.4 L) = 0.893 mol.
Next, let's calculate the number of moles of oxygen:
Molar mass of oxygen (O₂) = 2(16.00 g/mol) = 32.00 g/mol
55.7 L of oxygen at STP is equal to 55.7 L * (1 mol/22.4 L) = 2.48 mol.
Using the mole ratio from the balanced equation, we can calculate the moles of carbon dioxide produced:
(0.893 mol propane) * (3 mol carbon dioxide/1 mol propane) = 2.68 mol carbon dioxide
Finally, we can convert moles of carbon dioxide to volume using the ideal gas law at STP:
Volume of carbon dioxide = (2.68 mol carbon dioxide) * (22.4 L/mol) = 59.9 L
Therefore, 59.9 L of carbon dioxide gas is formed when 20 L of propane are reacted with 55.7 L of oxygen.
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.