Question

Liquid hexane

CH3CH24CH3

reacts with gaseous oxygen gas

O2

to produce gaseous carbon dioxide

CO2

and gaseous water

H2O

. If

4.93g

of carbon dioxide is produced from the reaction of

6.9g

of hexane and

6.6g

of oxygen gas, calculate the percent yield of carbon dioxide.

Round your answer to

2

significant figures.

Answer 1

Molar mass of C6H14,

MM = 6*MM(C) + 14*MM(H)

= 6*12.01 + 14*1.008

= 86.172 g/mol

mass(C6H14)= 6.9 g

use:

number of mol of C6H14,

n = mass of C6H14/molar mass of C6H14

=(6.9 g)/(86.17 g/mol)

= 8.007*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 6.6 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(6.6 g)/(32 g/mol)

= 0.2062 mol

Balanced chemical equation is:

2 C6H14 + 19 O2 ---> 12 CO2 + 14 H2O

2 mol of C6H14 reacts with 19 mol of O2

for 8.007*10^-2 mol of C6H14, 0.7607 mol of O2 is required

But we have 0.2062 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 = (12/19)* moles of O2

= (12/19)*0.2062

= 0.1303 mol

use:

mass of CO2 = number of mol * molar mass

= 0.1303*44.01

= 5.733 g

% yield = actual mass*100/theoretical mass

= 4.93*100/5.733

= 86%

Answer: 86%