Question

A gas tank in a home barbeque contains 13.2 kg of propane, C3H8. Calculate the heat (in kJ ) associated with the complete combustion of all the propane in the tank. C 3 H 8(g)+5O 2(g)→3CO 2(g) +4H 2 O (g)ΔH x×n =−2044 kJ

Answer 1

To calculate the heat released by the combustion of 13.2 kg of propane, we convert the mass to moles and then multiply by the heat of combustion per mole. This yields a total of 611,825.2 kJ of heat released during the complete combustion of the propane in the tank.

Step-by-step explanation:

To calculate the heat associated with the complete combustion of all the propane in the tank, we will use the heat of combustion of propane which is given as − 2044 kJ/mol. First, we need to find the number of moles of propane in 13.2 kg of propane.

Propane has a molecular weight of
`C_(3) H_(8)` which is 3(12.01) + 8(1.008) = 44.096 g/mol. To convert kilograms to grams, multiply by 1000, so we have 13.2 * 1000 = 13200 g of propane. The number of moles of propane is then 13200 g ÷ 44.096 g/mol = 299.3 moles of propane.

The total heat of combustion can be calculated by multiplying the number of moles of propane by the heat of combustion per mole:

Total heat = 299.3 moles * (− 2044 kJ/mol) = − 611,825.2 kJ

Since the heat of combustion is exothermic, it is released, and we can drop the negative sign to express that 611,825.2 kJ of heat is released.

Answer 2

To calculate the heat produced by burning 13.2 kg of propane, we convert the mass to moles and multiply it by the heat of combustion, resulting in a total heat release of − 611.6 MJ.

Step-by-step explanation:

To calculate the heat associated with the complete combustion of 13.2 kg of propane (C3H8), we first need to use the balanced equation for the combustion:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

The given heat of combustion for propane is − 2044 kJ per mole. First, we convert the mass of propane to moles by dividing it by the molar mass of propane:

(13.2 kg) / (44.09 g/mol) = 299.3 mol (noting that 1 kg = 1000 g)

Then, we multiply the number of moles by the heat of combustion:

299.3 mol × − 2044 kJ/mol = − 611,644.2 kJ

Therefore, the total heat associated with the combustion of 13.2 kg of propane is − 611.6 MJ (since 1 kJ = 0.001 MJ).