How much heat energy is obtained when 1 kg of ethane gas, C2H6, is burned in oxygen according to the equation: 2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)ΔHm= –3120 kJ(3.8.3)

52,000 kJ

Explanation:

The combustion reaction is the following:

2 C₂H₆(g) + 7 O₂(g) → 4CO₂(g) + 6H₂O(l)  ΔHm= –3120 kJ

From the chemical reaction and the value of ΔH we can see that 2 moles of ethane gas (C₂H₆(g)) release 3120 kJ of heat energy. So, we have the conversion factor: 3120 kJ/2 moles C₂H₆.

To calculate how many kJ are released from the combustion of 1 kg of ethane, we have to first convert the mass to moles by using the molecular weight (MW) of ethane (C₂H₆):

MW(C₂H₆) = (12 g/mol x 2 C) + (1 g/mol x 6 H) = 30 g/mol

mass = 1 kg x 1000 g/1 kg = 1000 g

moles of C₂H₆= mass/MW = 1000 g/(30 g/mol) = 33.3 mol C₂H₆

Finally, we multiply the conversion factor by the number of moles of C₂H₆ in 1 kg:

heat energy released = 33.3 mol C₂H₆ x 3120 kJ/2 moles C₂H₆ = 52,000 kJ

Therefore, 51,000 kJ are released when 1 kg of ethane is burned.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3