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alukav5142 [94]

2 years ago

10

A mixture of 620 lbmol/h of 36 mol% benzene and the remainder toluene is separated continuously into two products. One product i

s 98 mol% benzene and the other is 95 mol% toluene. Calculate the flow rates of both products in kmol/h.

1 answer:

Licemer1 [7]2 years ago

3 0

Answer:

$P_1=206.7lbmol/h\\\\P_2=413.3lbmol/h$

Explanation:

Hello!

In this case, for the separation of the benzene-toluene mixture, we can use the following mole balances including the given mole fraction at each stream per species:

$B:0.36*620=0.98P_1+0.05P_2\\\\T:0.64*620=0.02P_1+0.95P_2$

That can be solved by using a solver for P1 (benzene-rich flow) and P2 (toluene-rich flow):

$P_1=206.7lbmol/h\\\\P_2=413.3lbmol/h$

Best regards!

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The moles can be defined as the mass of the substance with respect to molar mass. The moles of potassium nitrate is 1 mol.

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The moles of a compound can be calculated from:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

The molarity can be defined as the moles of solute in a liter of solution.

The molarity can be expressed as:

$\rm Molarity=\dfrac{Moles\;\times\;1000}{Volume\;(mL)}$

The molarity of potassium nitrate solution is 2 M, and the volume is 500 mL.

The moles of potassium nitrate is given as:

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brainly.com/question/15209553

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1 year ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

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<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

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$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

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Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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