How many grams of sodium chloride could be produced from 41.3 L of chlorine gas?
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Answer:
The total amount of heat needed = 72.2116 kJ
Explanation:
Given that ;
the mass of acetic acid = 120.0 g
The initial temperature = 16.7 °C = (16.7 + 273.15 ) K = 298.85 K
The standard molar mass of acetic acid = 60.052 g/mol
Thus ; we can determine the number of moles of acetic acid;
number of moles of acetic acid = mass of acetic acid/ molar mass of acetic acid
number of moles of acetic acid = 120.0 g/ 60.052 g/mol
number of moles of acetic acid = 1.998 moles
For acetic acid:
The standard boiling point = 118.1 °C = ( 118.1 + 273.15 ) K = 391.25 K
The enthalpy of vaporization of acetic acid = 23.7 kJ/mol
The heat capacity of acetic acid c = 2.043 J/g.K
The change in temperature Δ T =
Δ T = (391.25 - 289.85)K
Δ T = 101.4 K
The amount of heat needed to bring the liquid acetic acid at 16.7°C to its boiling point is ;
q = mcΔT
From our values above;
q = 120 g × 2.043 J/g.K × 101.4 K
q = 24859.2 J
q = 24859 /1000 kJ
q = 24.859 kJ
we have earlier calculated our number of moles o f acetic acid to be 1.998 moles;
Thus;
The needed amount of heat =
The needed amount of heat =
The needed amount of heat = 47.3526 kJ
Hence;
The total amount of heat needed = 24.859 kJ + 47.3526 kJ
The total amount of heat needed = 72.2116 kJ