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MexChemE

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## Homework Statement

A certain pyrite ore contains 85% of FeS

_{2}and 15% of inerts. This ore is introduced into a roasting furnace with 20% excess air, in order to oxidize the FeS

_{2}in the reaction:

[tex]\textrm{FeS}_2 + \frac{11}{4}\textrm{O}_2 \rightarrow \frac{1}{2}\textrm{Fe}_2 \textrm{O}_3 + 2\textrm{SO}_2[/tex]

The solid product contains 2% FeS

_{2}in mass. Using 100 lb as basis, determine:

a) The chemical equation describing the process using the calculation basis, in mass and mole.

b) Conversion percentage.

c) The volume of the exhaust gases at 300 °C and 1 atm.

d) The amount of solid product obtained.

e) The amount of sulfuric acid which can be formed from the exhaust gases.

## Homework Equations

Steady state mass balance with chemical reaction

[tex]\textrm{In} + \textrm{Generation} = \textrm{Out} + \textrm{Consumption}[/tex]

## The Attempt at a Solution

First, I sketched a diagram of the process, as you can see in the attachments. Now, here's my work.

Part a)

85 lb (0.708 lbmol) of FeS

_{2}are being fed into the furnace. So, here's the balanced equation in molar base:

[tex]0.708\textrm{FeS}_2 + 1.947\textrm{O}_2 \rightarrow 0.354\textrm{Fe}_2 \textrm{O}_3 + 1.416\textrm{SO}_2[/tex]

And mass base:

[tex]85\textrm{FeS}_2 + 62.3\textrm{O}_2 \rightarrow 56.6\textrm{Fe}_2 \textrm{O}_3 + 90.6\textrm{SO}_2[/tex]

Part b)

Now, for part b) I need to know how much FeS

_{2}reacted (in lb). We'll call this quantity "A." We now also know that the O

_{2}required is 1.947 lbmol. Assuming dry air (21% oxygen; 79% nitrogen), the moles of air required are 9.271 lbmol; and since we have a 20% of excess air, the air fed into the furnace is 11.125 lbmol, of which 2.336 lbmol are oxygen (74.76 lb). Now, we'll analyze the mass exchange occurring with the reaction. We know 85 lb of FeS

_{2}and 74.76 lb O

_{2}were fed into the furnace. If A lb of FeS

_{2}reacted, we will have an output of:

(85 - A) FeS

_{2}

(74.76 - 0.73A) O

_{2}

(0.67A) Fe

_{2}O

_{3}

(1.07A) SO

_{2}

We know M

_{3}is composed of 15 lb of inerts plus some amount of iron sulphide and ferric oxide.

[tex]M_3=15+(85-A) + 0.67A= 100-0.33A[/tex]

Now, if we perform a sulphide balance we have:

[tex]85 = A + 0.02M_3[/tex]

Now we have a 2x2 linear equation system which we cal solve in order to obtain M

_{3}and A.

A = 83.551 lb

M

_{3}= 72.428 lb

The percentage of conversion will be given by:

[tex]\% \textrm{Conversion} = \frac{83.551}{85} \times 100\% = 98.3\%[/tex]

Part c)

Now that we know A, we can calculate the amount of moles of exhaust gases, which is 10.616 lbmol. Assuming ideal beahvior, at 300 °C and 1 atm of pressure, the exhaust gases occupy a volume of 7997.3 ft

^{3}.

Part d)

The solid product is M

_{3}, which are 72.428 lb, with a composition of:

(15 lb) Inerts

(1.499 lb) FeS

_{2}

(55.979 lb) Fe

_{2}O

_{3}

Part e)

I have some doubts about my procedure for this part. We have 1.397 lbmol of SO

_{2}, of which only 0.8604 lbmol can react with the limited amount of O

_{2}moles in the exhaust gases (which are 0.4302 lbmol) to form 0.8604 lbmol of SO

_{3}. This amount of SO

_{3}will in turn react with a stoichiometric amount of water in order to produce 0.8604 lbmol of H

_{2}SO

_{4}.

Therefore, with an output of 1.397 lbmol of SO

_{2}and 0.4302 lbmol of O

_{2}as limiting reactant, 0.8604 lbmol of H

_{2}SO

_{4}can be produced.

My concerns are mostly focused on part e), but feel free to point out any inconsistencies you may find along the way. Thanks in advance for any input!