In a closed system, heat should be conserved which means that the heat produced in the calorimeter is equal to the heat released by the combustion reaction. We calculate as follows:
Heat of the combustion reaction = mC(T2-T1)
= 1 (1.50) (41-21)
= 30 kJ
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:

Learn more:
Answer:
Signals from the A. Nervous system make the skeletal muscles move.
21.4 g Al * (1 mol / 26.98 g ) * (2 mol Fe / 2 mol Al) = 0.793 mol Fe
91.3 g Fe2O3 * (1 mol / 159.69 g) * (2 mol Fe / 1 mol Fe2O3) = 1.14 mol Fe
0.793 mol Fe * (55.85 g / 1 mol) = 44.3 g Fe produced.
In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=

%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%