Question

The enthalpy of combustion of ethane gas, C2H6(g), is about -1.5*103 kJ/mol. When ethane reacts with O2(g), the products are carbon dioxide CO2(g) and water H2O(l). How much heat is released during the combustion of ethane gas when 14 mols of O2(g) are consumed. Express your answer in kJ.

Answer 1

6.0×10^3 KJmol-1

Step-by-step explanation:

Equation of the reaction

C2H6(g) + 7/2 O2(g) ------> 2CO2(g) + 3H2O(g)

From the balanced reaction equation:

1 mole of ethane reacts with 3.5 moles of oxygen

x moles of ethane will react with 14 moles of oxygen

x= 14/3.5 = 4 moles of ethane

If heat of combustion of ethane= -1.5*103 kJ/mol

Then for 4 moles of ethane= 4× -1.5*103 kJ/mol= 6.0×10^3 KJmol-1

Answer 2

`6.0* 10^3 kJ` heat is released during the combustion of ethane gas when 14 moles of oxygen are consumed.

Step-by-step explanation:

`C_2H_6(g)+(7)/(2)O_2(g)\rightarrow 2CO_2(g)+3H_2O(l),\Delta H_(rxn)=-1.5* 10^3 kJ/mol`

According to reaction, when 7/2 moles of oxygen is consumed by 1 mol of ethane
`1.5* 10^3 kJ` energy is released.

Energy released when 1 mole of oxygen are consumed:
`(1.5* 10^3 kJ* 2)/(7)`

Then energy released when 14 moles of oxygen are consumed:

`(1.5* 10^3 kJ* 2)/(7)* 14=6.0* 10^3 kJ`

`6.0* 10^3 kJ` heat is released during the combustion of ethane gas when 14 moles of oxygen are consumed.