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.
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?
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