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dlinn [17]
3 years ago
5

What volume of 1.60 M hydrochloric acid solution (HCI) is required to neutralize 48.0 mL of 0.200 M sodium

Chemistry
1 answer:
nexus9112 [7]3 years ago
5 0

Answer:

Volume of HCl require =  6 mL

Explanation:

Given data:

Volume of HCl require = ?

Molarity of HCl solution = 1.60 M

Volume of NaOH = 48.0 mL

Molarity of NaOH = 0.200 M

Solution:

Formula:

M₁V₁     =      M₂V₂

By putting values,

1.60 M×V₁  =  0.200 M×48.0 mL

V₁  =  0.200 M×48.0 mL/1.60 M

V₁  =  9.6 M .mL /1.60 M

V₁  =  6 mL

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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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Explanation:

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