Answer the following questions. The combustion reaction of propene (C_3H_6) show

Question

Answer the following questions. The combustion reaction of propene (C_3H_6) shown below may be needed. 2 C_3H_6(g) + 9 O_2(g) rightarrow 6 CO_2(g) + 6 H_2O(g) A rigid 15.00 L cylinder is filled with propene gas at 22.5 degree C until the pressure is 6.67 atm. How many moles of propane gas is in the cylinder? If the temperature of the cylinder is raised to 65.3 degree C, what will the new pressure inside the cylinder be? If 5.50 moles of oxygen gas is added to the cylinder, how many moles of CO_2(g) can be made?

Explanation / Answer

Volume of rigid cylinder = 15L, T= 22.5+273= 295.5K, P= 6.67 atm, R =0.0821 L.atm/mole.K

  Number of moles, n = PV/RT =6.67*15/(0.0821*295.) =4.123 moles

When temperature is raised, P1/T1= P2/T2

=6.67/295.5 = P2/(65.3+273)= P2/338.3

P2= 338.3*6.67/295.5= 7.63 atm

3. The reaction of combustion is 2C3H6 + 9O2--à 6CO2 + 6H2O

Molar ratio (theoretical) of C3H6: O2= 2:9 =1:4.5

Actual ratio = 4.123 : 5 = 1: 5/4.123 =1:1.121

Oxygen is the limiting reactants, since oxygen required is 4.5 for 1 mole of C3H6. But supplied is only 1.121

9 moles of oxygen gives 6 moles CO2.

5 moles gives 5* 6/9 =3.33 moles of CO2.

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