Question

COLLEGE CHEMISTRY 35 POINTS

Determine the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen according to the following balanced equation:

2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(l)

Please show your work.

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.

Step-by-step 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 =
`(mass)/(atomic mass of 1 mole)`

number of moles=
`(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

`(16)/(2)` =
`(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 =
`(actual yield)/(theoretical yield)` x 100

percent yield =
`(12)/(176.04)` x 100

percent yield = 6.81 %