Question

The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?

Answer 1

C + O2= CO2
$n = \frac{m}{mw}$
$n = \frac{5.4}{12} \\ n = 0.45 \: mol \: of \: carbon$
$n = \frac{13.6}{12 + 16 \times 2} \\ n = \frac{13.6}{44} \\ n = 0.31 \: mol \: of \: carbon \: dioxide$
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %