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VladimirAG [237]
2 years ago
9

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)→4

CO2(g)+6H2O(l)ΔHm=–3120 kJ(3.8.3)
Chemistry
1 answer:
Xelga [282]2 years ago
5 0

Answer:

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.

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390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

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