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Marizza181 [45]
3 years ago
10

Attempt 3 During an experiment, a student adds 2.90 g CaO 2.90 g CaO to 400.0 mL 400.0 mL of 1.500 M HCl 1.500 M HCl . The stude

nt observes a temperature increase of 6.00 °C 6.00 °C . Assuming that the solution's final volume is 400.0 mL 400.0 mL , the density is 1.00 g / mL 1.00 g/mL , and the heat capacity is 4.184 J / g ⋅ °C 4.184 J/g⋅°C , calculate the heat of the reaction, Δ H rxn ΔHrxn . CaO ( s ) + 2 H + ( aq ) ⟶ Ca 2 + ( aq ) + H 2 O ( l )
Chemistry
1 answer:
kondor19780726 [428]3 years ago
3 0

Answer:

Explanation:

Equation of the reaction:

CaO(s) + 2H+(aq) -----> Ca2+(aq) + H2O(g)

The ∆Hrxn would be for one mole of CaO reacted or 2 moles of H+, whichever is the limiting reactant.

Number of moles = mass ÷ molar mass

Molar mass of CaO = 40 + 16

= 56 g/mol

moles of CaO = 2.90/56

= 0.0518 mol

Number of moles = concentration × volume

moles of HCl = 400 × 10^-3 × 1.500 = 0.6 moles

Moles of HCl = moles of H+

From the equation, 1 mole of CaO reacted with 2 moles of H+ to give 1 mole of water.

To find the limiting reagent,

0.6 mole of H+/2 moles of H+ × 1 mole of CaO

= 0.3 moles of CaO(> 0.0518 moles)

So, CaO is limiting reactant.

∆H = m × Cp × ∆T

m = density × volume

= 400 × 1

= 400 g

Cp = 4.184 J/g-ºC

∆T = +6 ºC

∆H = 400 × 4.184 × 6

= 10041.6 J

Since the reaction is exothermic,

∆Hrxn = -∆H/mol(CaO)

= -10041.6/0.0518

= -193853 J

= -193.9 kJ/mol.

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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:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

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 y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

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:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

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