A 9.780 g gascous mixture contains C_2H_6 and C_3H_8. Complete combustion of thi

Question

A 9.780 g gascous mixture contains C_2H_6 and C_3H_8. Complete combustion of this mixture requires 1.120 moles of oxygen gas. Calculate the mass percent of C_2H_6 in the original mixture What are the products of the complete combustion of a hydrocarbon C_xH^y? Which of the following applies: a single reaction, or parallel reactions? If parallel reactions are involved, can they lie added together in doing stoichiometric calculations? What can you define as unknown variables? How will they be used to determine the mass percent of C_2H_6? What equations represent (a) the total number of moles of oxygen reacted, and (b) the total mass of C_2H_6 and C_3H_8 in the original mixture, in terms of your defined unknown?

Explanation / Answer

C2H6 + 3.5O2      = 2CO2 + 3H2O      

C3H8 +   5O2       = 3CO2   + 4H2O   

2. parallel reactions; can be added in a defined proportion/ratio

3.

p, mass fraction of C2H6, the unknown variable

(1-p) , mass fraction of C3H8

C2H6 + 3.5O2      = 2CO2 + 3H2O       

C3H8 +   5O2       = 3CO2   + 4H2O   

                                               

Molar masses:   C2H6 30.07 g/mol ; C3H8 44.1 g/mol

C2H6 mol number = p · 9.780 g / 30.07 g/mol = 0.3252p

…has reacted with 3.5x0.3252p = 1.138p mol O2

C3H8 mol number = (1-p) · 9.780 g / 44.1 g/mol = 0.2218 (1-p) = 0.2218-0.2218p

..has reacted with 5(22.18-0.2218p) = 1.109 – 1.109p mol O2

4.

Total number of O2 moles :

            1.138p + 1.109 – 1.109p = 1.120 mol

            1.109 + 0.029p   mol = 1.120 mol

p = 0.379

p% = 37.9 %

For the masses:

   9.780 = p · 9.780 + (1-p) · 9.780

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