22.3 is the answer to this question
Answer:
Ethanamine (also known as ethylamine)
Explanation:
The compound that is requested by the question is ethanamine. Its trivial name is ethylamine.
It is a compound that contained the ethyl moiety (CH3CH2-) as well as the amine moiety (-NH2).
Ethanamine has a structure that can easily be determined by the statements in the question.
The structure of ethanamine is shown in the image attached.
Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
CO₂ + 4 H₄ → CH₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CO₂: 1 mole
- H₄: 4 moles
- CH₄: 1 mole
- H₂O: 2 moles
<h3>Moles of CH₄ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

<u><em>moles of CH₄= 340.4 moles</em></u>
Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
Learn more about the reaction stoichiometry:
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This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C