calculate the amount of water (H2O) in gram produced by the combustion of 80 g of Ethane
Answers
For methane
Combustion of methane :
CH 4 + 2O2 ⟶ CO2 +2H2O
Given weight of methane =16g
Molecular weight of methane =16g
No. of moles of methane in 16g of it =
mol. wt.
given wt.
=16/16
=1 mole
From the reaction,
Amount of water produced by the combustion of 1 mole of methane = 2 moles
weight of 1 mole of water =18g
⇒ weight of 2 moles of water =2×18=36g
Since, the given weight of methane is 16 g, hence the water produced by the combustion of 16 g of methane is 36 g.
For Ethane
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).