What is the percent yield if the reaction of 73.8 grams of O2 produces 25.6 grams of H2O?
2C3H8O2S+11O2-->6CO2+8H2O+2SO2
we first find the theoretical yield.
29.88
We then plug this into the equation (Actual Yield)/(Theoretical Yield) x 100
(25.6/29.88)= 0.856 mol x100
85.6%
That result is our answer.
2C3H8O2S+11O2-->6CO2+8H2O+2SO2
we first find the theoretical yield.
29.88
We then plug this into the equation (Actual Yield)/(Theoretical Yield) x 100
(25.6/29.88)= 0.856 mol x100
85.6%
That result is our answer.