Question

What is the percent yield of CO₂(𝑔) if 46.5 grams of CO₂(𝑔) is recovered from the reaction of 21.5 grams of C₂H₂(𝑔) according to the reaction below? 

2C2H2(g)+5 O2(g)—> 4CO2(g)+2H2O(g)

Answer 1

Moles of C₂H₂(𝑔) = $\frac{21.5}{26}=0.83 moles$

Moles of CO₂(𝑔) = $\frac{4×0.83}{2}=1.66moles$

Mass of CO₂= $1.66× 44=73.04moles$

Percent yield = $\frac{46.5}{73.04}×100=63.67$ %