1. -42.35 kJ
2. -169.4 kJ
3. -50.82 kJ
4. -236.7 kJ

Solution: -169.4 kJ

Given mass = 0.06 kg = 60 g

Molar mass of ethane = 2(12) + 6(1) = 30 g

number of moles of ethane in 60 grams = 60/30 = 2

We have the enthalpy of formation of 1 mol of ethane = -84.7 KJ/mol

So, for 60 g (2 moles) of ethane, enthalpy of formation = 2( -84.7) KJ/mol = -169.4 kJ