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Answer:
The total heat associated is -30,520.3 J.
Explanation:
Moles of ethanol = 0.499 moles
Molar mass of ethanol = 46 g/mol
Mass = Moles × Molar mass = 0.499 moles × 46 g/mol = 22.954 g
m is the mass of ethanol = 22.954 g
Q₁ is heat involved in the conversion of 22.954 g ethanol gas at 301°C to ethanol gas at 78.5°C
Thus, Q₁ = m × c × ΔT
Where,
c = The specific heat of the gas = 1.43 J/g°C
ΔT = Final temperature - Initial temperature = 78.5 - 301°C
= - 222.5 °C
Applying the values in the above equation as:-
Q₂ is the enthalpy of condensation from gas to liquid for the given mass of ethanol .
Thus, Q₂ = moles×ΔH condensation
Given that:- ΔH vaporization = 40.5 kJ/mol
Enthalpy of condensation of gaseous ethanol to liquid ethanol = - 40.5 kJ/mol
Considering, 1 kJ = 1000 J
So,
ΔH condensation = - 40.5 ×1000 J/mol = - 40500 J/mol
Thus, Q₂ = 0.499 moles × (- 40500 J/mol) = -20209.5 J
Q₃ is heat involved in the conversion of 22.954 g gaseous ethanol at 78.5°C to ethanol liquid at 25.0°C.
Thus, Q₃ = m × C ×ΔT
Where,
C = The specific heat of the liquid = 2.45 J/g°C
ΔT = Final temperature - Initial temperature = 25.0 - 78.5 °C
= - 53.5 °C
Applying the values in the above equation as:-
Applying the values as:
Total heat =
= -7303.3 J - 20209.5 J - 3007.5 J
= -30,520.3 J
The total heat associated is -30,520.3 J.
- The presence of salts and acids accelerates corrosion by producing conductive solutions that make electron transfer easier
Answer:
The answer to your question is: 70.8%
Explanation:
Data
Al₂O₃ = 60 g
C = 30 g
CO = gas
Al = 22.5 g
MW Al₂O₃ = 102 g
MW C = 12 g
MW Al = 54 g
Reaction
Al₂O₃ + 3C ⇒ 3 CO + 2 Al
Limiting reactant
102 g of Al₂O₃ -------------- 54 g Al
60 g -------------- x
x = 31.8 g
36 g of C ------------------ 54 g of Al
30 g of C ------------------ x
x = 45 g of Al
Limiting reactant = Al₂O₃
Percent yield =
Percent yield = 70.75 %