75 g of ethenol reacted with 105 g of oxygen in an experimen

75 g of ethenol reacted with 105 g of oxygen in an experiment. 65 g of CO2 was obtained. what is the percent yield for this reaction?

Solution

Molar mass of C2H5OH,
MM = 2*MM(C) + 6*MM(H) + 1*MM(O)
= 2*12.01 + 6*1.008 + 1*16.0
= 46.068 g/mol


mass(C2H5OH)= 75.0 g

use:
number of mol of C2H5OH,
n = mass of C2H5OH/molar mass of C2H5OH
=(75.0 g)/(46.068 g/mol)
= 1.628 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 105.0 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(105.0 g)/(32 g/mol)
= 3.281 mol
Balanced chemical equation is:
C2H5OH + 3 O2 ---> 2 CO2 + 3 H2O


1 mol of C2H5OH reacts with 3 mol of O2
for 1.628 mol of C2H5OH, 4.884 mol of O2 is required
But we have 3.281 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 = (2/3)* moles of O2
= (2/3)*3.281
= 2.188 mol


use:
mass of CO2 = number of mol * molar mass
= 2.188*44.01
= 96.27 g

% yield = actual mass*100/theoretical mass
= 65*100/96.27
= 67.5 %
Answer: 67.5 %