10.0cm of propane was sparked with 60.00 of oxygen at S.T.P what is the volume of the residual gas after reaction is cooled
what is the reaction equation?
how many moles of gas produced?
each mole is 22.4L
To find the volume of the residual gas after the reaction is cooled, we first need to determine the balanced chemical equation:
C3H8 + 5O2 → 3CO2 + 4H2O
According to the balanced equation, 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.
Given that the initial volume of propane is 10.0 cm^3 and the volume of oxygen is 60.00 cm^3, we can convert these volumes to moles using the ideal gas law equation:
PV = nRT
At standard temperature and pressure (STP):
P = 1 atm
T = 273.15 K
R = 0.0821 atm·L/mol·K
For propane:
V_propane = 10.0 cm^3 = 0.0100 L
For oxygen:
V_oxygen = 60.00 cm^3 = 0.0600 L
Now we can calculate the number of moles of propane and oxygen:
n_propane = (P * V_propane) / (R * T)
n_oxygen = (P * V_oxygen) / (R * T)
n_propane = (1 atm * 0.0100 L) / (0.0821 atm·L/mol·K * 273.15 K)
n_oxygen = (1 atm * 0.0600 L) / (0.0821 atm·L/mol·K * 273.15 K)
n_propane ≈ 0.00044 mol
n_oxygen ≈ 0.0033 mol
Since the stoichiometric ratio between propane and oxygen is 1:5, we can determine that 0.00044 mol of propane reacts with 5 * 0.00044 mol of oxygen.
n_carbon_dioxide = 3 * 0.00044 mol
n_water = 4 * 0.00044 mol
To find the total moles of the residual gas, we subtract the moles of carbon dioxide and water formed from the total moles of propane and oxygen consumed:
n_residual_gas = (n_propane + n_oxygen) - (n_carbon_dioxide + n_water)
n_residual_gas = (0.00044 mol + 0.0033 mol) - (3 * 0.00044 mol + 4 * 0.00044 mol)
n_residual_gas = 0.00220 mol - 0.00264 mol
n_residual_gas = -0.00044 mol (negative because there is an excess of reactants)
Since we have an excess of reactants, the residual gas will occupy the same volume as the initial volume of the propane:
V_residual_gas = V_propane
V_residual_gas = 10.0 cm^3
Therefore, the volume of the residual gas after the reaction is cooled is 10.0 cm^3.
To find the volume of the residual gas after the reaction is cooled, we need to determine the stoichiometry of the reaction first. The balanced equation for the combustion of propane (C₃H₈) with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) is:
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 and 4 moles of water.
Given that the volume of propane is 10.0 cm³ and the volume of oxygen is 60.00 cm³ at STP (standard temperature and pressure), we can use the ideal gas law to calculate the number of moles of each gas:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume (converted to liters, so 10.0 cm³ = 0.01 L for propane and 60.00 cm³ = 0.06 L for oxygen)
n = number of moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (STP = 273 K)
For propane:
n₁ = (P₁ * V₁) / (R * T) = (1 atm * 0.01 L) / (0.0821 L·atm/mol·K * 273 K)
For oxygen:
n₂ = (P₂ * V₂) / (R * T) = (1 atm * 0.06 L) / (0.0821 L·atm/mol·K * 273 K)
Now, we need to determine which reactant is limiting. Since the ratio of propane to oxygen in the balanced equation is 1:5, we can calculate the amount of propane needed to react with the given amount of oxygen:
propane needed = (5/1) * n₂ = 5 * n₂
If the number of moles of propane (n₁) is greater than the propane needed, then oxygen is limiting. Otherwise, propane is limiting.
Once we determine the limiting reactant, we can calculate the number of moles of carbon dioxide and water produced based on the stoichiometry of the balanced equation.
Finally, we can use the ideal gas law again to find the volume of the residual gas (carbon dioxide and water) after cooling, assuming the reaction goes to completion and the gases are at STP conditions.
I hope this explanation helps you understand how to approach the problem step by step.