Question

An excess of methane reacts with 10.5 g of oxygen gas, O2 in the following reaction.


CH4(g)+2O2(g)→CO2(g)+2H2O(g)



(a) Calculate the mass of carbon dioxide gas, CO2 produced and
(b) the percentage yield if only 3.8 g of CO2 is actually isolated.

Answer 1

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
mols O2 = g/molar mass = 10.5/32 = estimated 0.3
Convert mols O2 mols CO2 using the coefficients in the balanced equation like this. 0.3 mols O2 x (1 mols CO2/2 mols O2) = estimated 0.3/2 or about 0.15
Then grams CO2 = mols CO2 x molar mass CO2 = 0.15 x 44 = about 7 but you need to redo all the calculations. This is the theoreticfal yield (TY).
b. The actual yield (AY) is 3.8 g.
%yield = (AY/TY)*100 = ?