There are 1825.6 g in a 14.5 moles Lithium permanganate
<h3> Further explanation</h3>The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
Moles can also be determined from the amount of substance mass and its molar mass :
moles of Lithium permanganate = 14.5
Lithium permanganate (LiMnO4) MW=125.9 g/mol, so mass :
This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and and
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
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