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Sveta_85 [38]
3 years ago
13

How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol CsH18? C8H18 +

25/2 O2 O 8CO2 + 9H20 e. 5500 kJ a. 7200 kJ b. 8360 kJ c. 4.1 x 105 kJ d. 3600 kJ
Chemistry
1 answer:
aalyn [17]3 years ago
5 0

Answer : The correct option is, (D) 3600 kJ

Explanation :

Mass of octane = 75 g

Molar mass of octane = 114.23 g/mole

Enthalpy of combustion = -5500 kJ/mol

First we have to calculate the moles of octane.

\text{ Moles of octane}=\frac{\text{ Mass of octane}}{\text{ Molar mass of octane}}=\frac{75g}{114.23g/mole}=0.656moles

Now we have to calculate the heat released in the reaction.

As, 1 mole of octane released heat = -5500 kJ

So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)

                                                                   = -3608 kJ

                                                                   ≈ -3600 kJ

Therefore, the heat released in the reaction is 3600 kJ

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solniwko [45]

Answer:

1. Volume as STP = 755 L

2. Outside temperature = 255 K

3. Percentage yield = 70.5%

Explanation:

1. At STP, pressure  = 101.3 kpa, temperature  = 0°C or 273.15 K

Using the general gas equation :

P1V1/T1 = P2V2/T2

P1 = 620 kpa

V1 = 140 L

T1 = 37°C or (273.15 + 37) K = 310.15 K

P2 = 101.3 kpa

V2 = ?

T2 = 273.15 K

V2 = P1V1T2/P2T1

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V1 = 2.5 L

T1 = 290 K

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T2 = 2.2 × 290 / 2.5

T2 = 255 K

3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu

From equation of the reaction,  2 moles of Al produces 3 moles of Cu

Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g

2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g

54 g of Al produces 190.5 g of Cu

1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu

Percentage yield = actual yield /theoretical yield × 100%

Percentage yield = 4.65/6.60 × 100%

Percentage yield = 70.5%

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3 years ago
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Answer:

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