Amount of water producer in a gram produced by the combustion of 8 gra

1.	Amount of water producer in a gram produced by the combustion of 8 gram of ethane
Answer» Answer: FIRST, I would WRITE a balanced equation for the combustion reaction, using 1 mol of ethane: C2H6(g) + 3.5O2(g) = 2CO2(g) + 3H2O(g) T = 2950C But 8g of ethane C2H6 is 8/30 = 0.267mol, so the equation is (multiplying the above equation by 0.267: 0.267C2H6(g) + 0.9345O2(g) = 0.534CO2(g) + 0.801H2O(g) T = 2950C Change in FREE Energy: ΔG(2950C) = -420.4kJ (negative, so the reaction RUNS) Change in Enthalpy: ΔH(2950C) = -389.5kJ (negative, so the reaction is EXOTHERMIC) This reaction produces 0.801mol of water, which is 14.43g, 17.953L (as a gas)

Amount of water producer in a gram produced by the combustion of 8 gram of ethane

Answer»

Answer:

FIRST, I would WRITE a balanced equation for the combustion reaction, using 1 mol of ethane:

C2H6(g) + 3.5O2(g) = 2CO2(g) + 3H2O(g) T = 2950C

But 8g of ethane C2H6 is 8/30 = 0.267mol, so the equation is (multiplying the above equation by 0.267:

0.267C2H6(g) + 0.9345O2(g) = 0.534CO2(g) + 0.801H2O(g) T = 2950C

Change in FREE Energy: ΔG(2950C) = -420.4kJ (negative, so the reaction RUNS)

Change in Enthalpy: ΔH(2950C) = -389.5kJ (negative, so the reaction is EXOTHERMIC)

This reaction produces 0.801mol of water, which is 14.43g, 17.953L (as a gas)