Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol
9.
(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3
10.
Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:
NaHCO3 + H{+} = Na{+} + H2O + CO2
(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory
11.
n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2
12.
(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%
Explanation:
Answer:
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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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