Question

The reaction of 8.00 g of carbon with excess o2 yields 10.9 g of co2. what is the percent yield of this reaction?

Answer 1

C+O2----> CO2
8.00g              10.9 g

8.00g C*1 molC/12.0g C= 0.667 mol C
10.2g CO2*1mol/(12.0+2*16.0)=0.232 mol CO2  actual yield
0.667 mol C gives 0.667 mol CO2 theoretical yield
% yield = (actual yield/theoretical yield)*100%=
=(0.232 mol CO2/0.667 mol CO2)*100%= 34.9%