2C2H6(g)+7O2(g)--> 4Co2 (g)+6H2O(l)

Question

A) if 3.00 moles of C2H6 and 9.00 moles of O2 are introducedinto an empty container at 32.0 degree C and then ignited tointiate the above reaction calculate the mass of water that isproduced.
B) if the container volume is 675 what mass of water is in thevapour phase?

Answer 1

Convert 3.00 C2H6 to mols H2O. That's 3.00 x (6 mol H2O/2 mols C2H6) = 9.00 mols H2O

Do the same for mols O2 to mols H2O. That's 9.00 x (6 mols H2O/7 mols O2) = 7.71 mols H2O.

In limiting reagent problems the smaller number is ALWAyS the correct choice; therefore, O2 is the limiting reagent and 7.71 mols H2O will be formed. Then g H2O = mols H2O x molar mass H2O = ?

B. That's 675 what? mL? L?
Look up the vapor pressure H2O @ 32 C, plug all of the numbers into PV = nRT and solve for n. Then n = grams/molar mass. You know n and molar mass, solve for mass H2O in the vapor phase.