C3H8 (g) + 5O2 (g) <-----> 3CO2 + 4HOH
If such an oxidation is carried out on a gaseous mixture prepared by mixing 6.1 grams of propane and 11.0 L of oxygen measured at 1.0atm and 150 degrees celcius, what will be the total final volume under the same conditions of measurement after the reaction is complete?
First, note the correct spelling of celsius. Second, is the CO2 a gas and is the water a gas? I assume so at 150 C but you don't have it in the equation.
This is a limiting reagent problem. You know that because amounts for both reactants are given. You must determine which is the limiting reagent.
1. Convert 6.1 g C3H8 to moles. moles = grams/molar mass. Then use PV = nRT to convert to volume at the conditions listed.
2a. Using the coefficients in the balanced equation, convert L C3H8 to L CO2. Same procedure for L H2O.
2b. Same procedure, convert L O2 to L CO2 and L H2O.
2c. You will have two sets of numbers (one using C3H8 and the other O2) and both can't be correct; the correct set will be the smaller set of values) and the reagent producing the smaller set will be the limiting reagent.
2d. Same procedure, convert L of the limiting reagent to L of the "other" reagent, the subtract from the initial volume to determine the amount of the "other" reagent that didn't react.
3. Now add the volume of the limiting reagent (which will be zero), + volume of the other reagent + volume CO2 + volume H2O to find final volume.
All of this may sound complicated but it isn't after you see how the solution is being conducted. Post your work if you get stuck and I can help you through it.
To solve this problem, we need to determine the number of moles of each gas involved in the reaction and use the stoichiometry of the balanced equation to find the volume of the final mixture.
1. Convert the mass of propane (C3H8) to moles:
- The molar mass of propane is 44.1 g/mol (3*12.01 + 8*1.01 = 44.1).
- Moles of propane = mass of propane / molar mass of propane.
Moles of propane = 6.1 g / 44.1 g/mol = 0.1387 mol.
2. Convert the volume of oxygen (O2) to moles:
- The reaction is carried out at 1.0 atm and 150 degrees Celsius, so we need to use the ideal gas law to convert the volume to moles.
- PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
- Convert 150 degrees Celsius to Kelvin: 150°C + 273.15 = 423.15 K.
- Rearrange the ideal gas law equation to solve for n: n = PV / RT.
- Moles of oxygen = (1.0 atm * 11.0 L) / (0.0821 L·atm/(mol·K) * 423.15 K).
Moles of oxygen = 0.2916 mol.
3. Determine the limiting reactant:
- The balanced equation tells us that propane (C3H8) and oxygen (O2) are in a 1:5 mole ratio.
- Since there is less propane than oxygen, propane is the limiting reactant.
- This means that propane will be completely consumed, and oxygen will be in excess.
4. Use the mole ratio to find the number of moles of carbon dioxide (CO2) and water (H2O) formed:
- From the balanced equation, the mole ratio of propane to carbon dioxide is 1:3, and the mole ratio of propane to water is 1:4.
- Moles of carbon dioxide = moles of propane * (3 moles of CO2 / 1 mole of propane).
Moles of carbon dioxide = 0.1387 mol * (3/1) = 0.4161 mol.
- Moles of water = moles of propane * (4 moles of H2O / 1 mole of propane).
Moles of water = 0.1387 mol * (4/1) = 0.5548 mol.
5. Calculate the total final moles of gas:
- The total final moles of gas will be the sum of the moles of carbon dioxide, water, and any remaining oxygen.
- Moles of oxygen remaining = moles of oxygen initially - moles of propane used.
Moles of oxygen remaining = 0.2916 mol - 0.1387 mol = 0.1529 mol.
- Total final moles of gas = moles of carbon dioxide + moles of water + moles of oxygen remaining.
Total final moles of gas = 0.4161 mol + 0.5548 mol + 0.1529 mol = 1.1238 mol.
6. Convert moles of gas to volume:
- Considering that the reaction is carried out under the same conditions of measurement, we can use the ideal gas law to calculate the final volume.
- Rearrange the ideal gas law equation to solve for V: V = nRT / P.
- Final volume = (1.1238 mol * 0.0821 L·atm/(mol·K) * 423.15 K) / 1.0 atm.
Final volume = 38.33 L.
Therefore, the total final volume of the gas mixture under the same conditions of measurement after the reaction is complete is 38.33 liters.