posted by Alicia on .
Assuming the volumes of all gases in the reaction are measured at the same temperature and pressure, calculate the volume of water vapor obtainable by the explosive reaction of a mixture of 725mL of hydrogen gas and 325mL of oxygen gas.
Calculate the volume of air at 30 degrees C and 1.00atm that is needed to burn completely 10.0 grams of propane. Assume that air is 21.0% oxygen (O2) by volume.
2H2 + O2 ==> 2H2O
This may be solved the usual stoichiometric way by changing g H2 and g O2 to mols, determining which is the limiting reagent, etc etc. OR, the shortcut way is to simply use volumes since all are gases. If you don't get this way, I can explain how it works in detail. Let me know.
To use volumes only, 725 mL H2 will form 725 mL H2O since the ratio of H2 to H2O is 2:2 (or 1:1). 325 mL O2 will form 325 x (2 mols H2O/1 mol O2) = 650 mL H2O. Therefore, 650 mL H2O will be the final product (O2 is the limiting reagent). There will be excess H2 remaining unreacted.
For #2, write the equation and balance it.
C3H8 + 5O2 ==> 3CO2 + 4H2O
Convert 10 g propane to mols. #mols = grams/molar mass propane.
Convert mols Propane to mols O2 using the coefficients in the balanced equation.
Use PV = nRT to convert mols O2 to volume O2 at the conditions cited.
Convert to volume of air by the following:
volume oxygen at cited conditions/0.21 = volume air.
Post your work if you get stuck or if you have any questions.
I have a problem that states:
Find the pressure of a sample of carbon tetrachloride, CCl4 if 1.00 mol occupies 35.0L at 77.0 degress C (slightly above its normal boiling point). Assume that CCl4 obeys (a) the ideal gas law; (b) the van der Waals equation. The van der Waals constants for CCl4 are a= 20.39(L^2)(atm)/mol^2 and b= 0.1383L/mol.
I tried to do the problems and the answers I got are (a)=.814atm and (b)=.800atm but the back of the book is telling me that (a)=0.821atm an (b)=0.805. I'm not sure if i'm wrong or if the book is off.