Question

When 85g of CH4 are mixed with 85g of O2 according to this equation:

CH4 + 2O2 - cO2 + 2H2O

What is the theoretical yield in gams of CO2?
What is the percent yield if 45.7g of CO2 are produced?

Please explain!

Answer 1

This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. This is worked just like two simple stoichiometry problems.

CH4 + 2O2 - cO2 + 2H2O

mols CH4 = grams/molar mass = ?
mols O2 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols CH4 to mols CO2.
Do the same to convert mols O2 t mnols CO2.
It is likely these two values for mols CO2 will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that number is the LR.
Using the smaller value, convert to g CO2. g = mols CO2 x molar mass CO2. This is the theoretical yield (TY). The actual yield (AY) is 45.7g.
%yield = (AY/TY)*100 = ?