what is the percent yield if 27 g of h20 obtained when 33 g of c3h8 reacts with 56 L of 02 at STP.The secong product of reaction is CO2
This is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
C3H8 + 5O2 ==> 3CO2 + 4H2O
mols C3H8 = grams/molar mass
mols O2 = L/22.4 = ?
Convert mols C3H8 to mols H2O
Convert mols O2 to mols H2O.
It is likely these two values will not be the same which means one of them must be wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Convert the smaller value to grams (g = mols x molar mass). This is the theoretical yield (TY)
The actual yield (AY) is 27g from the problem.
%yield = (AY/TY)*100 = ?