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timofeeve [1]
2 years ago
10

Calculate ΔH when 2 moles of ethanol (C2H5OH) reacts with excess oxygen according to the following thermochemical equation?C2H5O

H + 3O2 → 2CO2 + 3H2O ΔH = −1366.7 kJ
Chemistry
1 answer:
klio [65]2 years ago
5 0

Answer:

–2733.4 KJ

Explanation:

The balanced equation for the reaction is given below:

C₂H₅OH + 3O₂ —> 2CO₂ + 3H₂O

ΔH = −1366.7 kJ

From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Finally, we shall determine the enthalpy change (ΔH) produced by the reaction of 2 moles of C₂H₅OH. This can be obtained as follow:

From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Therefore, 2 moles of C₂H₅OH will react to produce enthalpy change (ΔH) of = 2 × −1366.7 = –2733.4 KJ.

Thus, enthalpy change (ΔH) obtained is –2733.4 KJ

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.400 moles of CO2 gas are confined in a 5.00-liter container at 25 °C. Calculate the pressure exerted in atmospheres and mm Hg.
solong [7]

The pressure exerted by 0.400 moles of carbon dioxide in a 5.00 Liter container at 25 °C would be 1.9563 atm or  1486.788 mm Hg.

<h3>The ideal gas law</h3>

According to the ideal gas law, the product of the pressure and volume of a gas is a constant.

This can be mathematically expressed as:

pv = nRT

Where:

p = pressure of the gas

v = volume

n = number of moles

R = Rydberg constant (0.08206 L•atm•mol-1K)

T = temperature.

In this case:

p is what we are looking for.

v = 5.00 L

n = 0.400 moles

T = 25 + 273

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Now, let's make p the subject of the formula of the equation.

p = nRT/v

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Recall that: 1 atm = 760 mm Hg

Thus:

1.9563 atm = 1.9563 x 760 mm Hg

                   = 1486.788 mm Hg

In other words, the pressure exerted by the gas in atm is 1.9563 atm and in mm HG is 1486.788 mm Hg.

More on the ideal gas law can be found here: brainly.com/question/28257995

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In a separate experiment beginning from the same initial conditions, including a temperature Ti of 25.0°C, half the number of mo
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Answer:

- 178 ºC

Explanation:

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PV = nRT,

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For the initial conditions :

P₁ V₁ = n₁ R T₁    (1)

and for the final conditions:

P₂V₂= n₂ R T₂    where   n₂ = n₁/2     then    P₂ V₂ = n₁/2 T₂    (2)

Assuming V₂ = V₁ and  dividing (2) by Eqn (1) :

P₂ V₂ = n₁/2 R T₂  / ( n₁ R T₁)      then  P₂ / P₁ = 1/2 T₂ / T₁

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T₂ = 95 - 273 = - 178 º C

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