6.81% is the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen.
Explanation:
Data given:
mass of carbon dioxide formed = 12 grams (actual yield)
atomic mass of CO2 = 44.01 grams/mole
moles of [tex]C_{8} H_{18}[/tex] = 0.5
Balanced chemical reaction:
2 [tex]C_{8} H_{18}[/tex] + 25 [tex]O_{2}[/tex] ⇒ 16 C[tex]O_{2}[/tex] + 18 [tex]H_{2} _O{}[/tex]
number of moles of carbon dioxide given is
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
number of moles= [tex]\frac{12}{44}[/tex]
number of moles of carbon dioxide gas = 0.27 moles
from the reaction 2 moles of [tex]C_{8} H_{18}[/tex] reacts to produce 16 moles of C[tex]O_{2}[/tex]
So, when 0.5 moles reacted it produces x moles
[tex]\frac{16}{2}[/tex] = [tex]\frac{x}{0.5}[/tex]
x = 4
4 moles of carbon dioxide formed, mass from it will give theoretical yield.
mass = number of moles x molar mass
mass = 4 x 44.01
= 176.04 grams
percent yield =[tex]\frac{actual yield}{theoretical yield}[/tex] x 100
percent yield = [tex]\frac{12}{176.04}[/tex] x 100
percent yield = 6.81 %
