Question

COLLEGE CHEMISTRY 35 POINTS

Answer 1

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 $C_{8} H_{18}$ = 0.5

Balanced chemical reaction:

2 $C_{8} H_{18}$ + 25 $O_{2}$ ⇒ 16 C$O_{2}$ + 18 $H_{2} _O{}$

number of moles of carbon dioxide given is

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles= $\frac{12}{44}$

number of moles of carbon dioxide gas = 0.27 moles

from the reaction 2 moles of $C_{8} H_{18}$ reacts to produce 16 moles of C$O_{2}$

So, when 0.5 moles reacted it produces x moles

$\frac{16}{2}$ = $\frac{x}{0.5}$

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 =$\frac{actual yield}{theoretical yield}$ x 100

percent yield = $\frac{12}{176.04}$ x 100

     percent yield  = 6.81 %