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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Answer:
Explanation:
Given parameters:
number of moles magnesium = 6.80mol
number of moles of chlorine = 13.56mol
To find the complete chemical formula, we should obtain the formula of the compound.
Mg Cl
Number of
moles 6.8 13.56
Dividing
by the smallest 6.8/6.8 13.56/6.8
1 2
The formula of the compound is MgCl₂
This is an ionic compound in which Magnesium loses two electrons that would be gained by Cl atoms requiring just an electron each to complete their octet.
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