Question

Liquid hexane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 6.55g of carbon dioxide is produced from the reaction of 3.45g of hexane and 17.1g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 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)= 3.45 g

use:

number of mol of C6H14,

n = mass of C6H14/molar mass of C6H14

=(3.45 g)/(86.17 g/mol)

= 4.004*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 17.1 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(17.1 g)/(32 g/mol)

= 0.5344 mol

Balanced chemical equation is:

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

2 mol of C6H14 reacts with 19 mol of O2

for 4.004*10^-2 mol of C6H14, 0.3803 mol of O2 is required

But we have 0.5344 mol of O2

so, C6H14 is limiting reagent

we will use C6H14 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 = (12/2)* moles of C6H14

= (12/2)*4.004*10^-2

= 0.2402 mol

use:

mass of CO2 = number of mol * molar mass

= 0.2402*44.01

= 10.57 g

% yield = actual mass*100/theoretical mass

= 6.55*100/10.57

= 62.0 %

Answer: 62.0 %