Question

7) Give the percent yield when 28.16 g of CO2 are formed from the reaction of 6.000 moles of C8H18 with 3.000 moles of 02 2 C8H18+25 02 16 CO2+ 18 H20 A) 33.34% B) 8.334% C) 13.33% D) 75.00 E) 16.67%

Answer 1

Balanced chemical equation is:

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

2 mol of C8H18 reacts with 25 mol of O2

for 6 mol of C8H18, 75 mol of O2 is required

But we have 3 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (16/25)* moles of O2

= (16/25)*3

= 1.92 mol

use:

mass of CO2 = number of mol * molar mass

= 1.92*44.01

= 84.5 g

% yield = actual mass*100/theoretical mass

= 28.16*100/84.5

= 33.34%

Answer: A