Asked by Anonymous
2C2H6(g)+7O2(g)--> 4Co2 (g)+6H2O(l)
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?
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?
Answers
Answered by
DrBob222
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.
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.
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