Question

If you add 179.0 mL of C3H8 (propane) gas at STP to a 1.00L container, seal the container, and combust the gas with 390mL of oxygen. (Molar volume of a gas at STP=22.4L/mol)
a) what is the balanced chem. equation for the reaction?
b) what is the limiting reagent? what are the theoretical moles of CO2 produced?
c) Use daltons law of partial pressure and the ideal gas law to determine the partial pressure of each of the gases when the reaction goes to completiom at 299.2K

Answer 1

C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g)

    1 mol C3H8 = 5 mol O2 = 3 mol CO2 = 4 mol H2O

NO of mol of propane burned = 0.179/22.4 = 0.008 mol

NO of mol of O2 added = 0.39/22.4 = 0.0174 mol

limiting reactant = O2

theoretical moles of CO2 formed = 0.0174*3/5 = 0.01044 mol

excess propane gas present = (0.008 - (0.0174*1/5))= 0.00452 mol

c)

   according to daltons law of partial pressure

TOTAL moles = 0.01044+0.00452 = 0.01496 mol

Ptotal = nRT/V

         = 0.01496*0.0821*273.15/1

          = 0.335 atm

pTotal = pC3H8 + pCO2

pCO2 = (0.01044/(0.01044+0.00452))*0.335

       = 0.234 atm

   pC3H8 = (0.00452/(0.01044+0.00452))*0.335

         = 0.1 atm