Question

A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.4 mole% water. After all the water is removed from the products, the residual gas contains 69.4 mole% CO2 and the balance O2.

a) Write the element balances in mass units

Answer 1

We have a mixture of propane and butane

The balance chemical equation for

butane is

C4H10 + 13/2 O2 ===> 4 CO2 + 5 H2O

for propane it is

C3H8 + 5O2 → 3CO2 + 4H2O

water is 47.4 mol% in the remaining 52.6% CO2 is 69.4% and O2 is 30.6%

So if I have 100 moles of total mixture 47.4 mol% is water 52.6 x 0.694 mol% is CO2 and 0.526 x 0.306 mol% is O2

47.4 x 18g/mol = 853 g of water,

52.6 x 0.694 = 36.5 mol% x 44 g/mol = 1606 g of CO2

52.6 x 0.306 x 32 g/mol = 515 g of O2

To find the ratio of propane and butane in the mixture we will need to combine the two

nC4H10 + 13/2 O2 ===> 4*n CO2 + 5*n H2O

mC3H8 + 5O2 → 3*mCO2 + 4*mH2O

4 *n CO2 + 3 * m CO2 = 36.5 CO2

5 *n H2O + 4 * m H2O = 47.4 H2O

m= 7.6 and n = 3.4

So Butane 3.4 moles and propane is 7.6 moles

So Butane is 3.4 x 58.12 g/mol = 199.06 g

Propane is 7.6 x 44.1 g/mol = 335.1 g