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Chapter 4: Problem 59

Titanium dioxide \(\left(\mathrm{Ti} \mathrm{O}_{2}\right)\) is used extensively as a white pigment. It is produced from an ore that contains ilmenite \(\left(\mathrm{FeTiO}_{3}\right)\) and ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right) .\) The ore is digested with an aqueous sulfuric acid solution to produce an aqueous solution of titanyl sulfate \(\left[(\mathrm{TiO}) \mathrm{SO}_{4}\right]\) and ferrous sulfate (FeSO \(_{4}\) ). Water is added to hydrolyze the titanyl sulfate to \(\mathrm{H}_{2} \mathrm{TiO}_{3},\) which precipitates, and \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) The precipitate is then roasted, driving off water and leaving a residue of pure titanium dioxide. (Several steps to remove iron from the intermediate solutions as iron sulfate have been omitted from this description.) Suppose an ore containing \(24.3 \%\) Ti by mass is digested with an \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) solution, supplied in \(50 \%\) excess of the amount needed to convert all the ilmenite to titanyl sulfate and all the ferric oxide to ferric sulfate \(\left[\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\right] .\) Further suppose that \(89 \%\) of the ilmenite actually decomposes. Calculate the masses (kg) of ore and 80\% sulfuric acid solution that must be fed to produce \(1000 \mathrm{kg}\) of pure \(\mathrm{TiO}_{2}\)

Short Answer

Expert verified
To produce 1000 kg of \(TiO_{2}\), you need approximately 4625 kg of ore and 8661 kg of 80% sulfuric acid solution.

Step by step solution

01

Calculate the mass of \(\mathrm{TiO}_{2}\) obtained from ilmenite

We're given that the ore is composed of 24.3% Ti by mass and there is 89% decomposition of ilmenite. So, the mass of \(\mathrm{TiO}_{2}\) obtained from 100% decomposition is \(0.243 \times 1000 = 243 \mathrm{kg}\). However, since only 89% decomposes, the actual yield of \(\mathrm{TiO}_{2}\) would be \(0.89 \times 243 = 216.27 \mathrm{kg}\).
02

Determine the mass of ore

To get 1000 kg of \(\mathrm{TiO}_{2}\), the required ore is calculated as \( \frac{1000 \mathrm{kg}}{216.27 \mathrm{kg}} \times 1000 \mathrm{kg} = 4624.57 \mathrm{kg}\). So, the amount of ore needed is approximately 4625 kg.
03

Determine the mass of sulfuric acid

Given the sulfuric acid solution is supplied in 50% excess of the amount needed, the amount of 80% sulfuric acid solution required is calculated as \( \frac{4624.57 \mathrm{kg}}{0.8} \times 1.5 = 8661.06 \mathrm{kg}\). So, approximately 8661 kg of 80% sulfuric acid solution is needed.
04

Conclusion

Therefore, to produce 1000 kg of titanium dioxide, approximately 4625 kg of the ore and 8661 kg of 80% sulfuric acid solution would need to be used.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Titanium Dioxide Production
Titanium dioxide, known by its chemical symbol \(\mathrm{TiO}_2\), is a versatile compound used in products ranging from paint to sunscreen. Its brilliance and opacity make it a sought-after white pigment. This compound is primarily produced through processes involving ilmenite \(\left(\mathrm{FeTiO}_3\right)\) and ferric oxide \(\left(\mathrm{Fe}_2\mathrm{O}_3\right)\).
The process begins with digesting the ore in an acid, producing titanyl sulfate \(\left((\mathrm{TiO}) \mathrm{SO}_4\right)\) and ferrous sulfate (\(\mathrm{FeSO}_4\)). Water is then introduced to trigger the hydrolysis of titanyl sulfate, forming a titanium hydroxide precipitate \(\mathrm{H}_2\mathrm{TiO}_3\), which is later turned into \(\mathrm{TiO}_2\) upon roasting. By removing excess water, pure titanium dioxide is obtained. Understanding this chain of chemical reactions is vital for efficient \(\mathrm{TiO}_2\) production.
This elaborative process ensures purity while maintaining the essential properties that make titanium dioxide a valuable industrial compound.
Stoichiometry
Stoichiometry is fundamental in chemical engineering and is especially crucial in calculating the quantities needed and produced in reactions. In our context, it involves determining how much ore and sulfuric acid is required to yield a target quantity of titanium dioxide.
In the problem, stoichiometry is applied by accounting for decomposition percentages and purity constraints. Approximately 24.3% of the ore consists of titanium, and out of this, 89% actually decomposes. Hence, for a 1000 kg target of \(\mathrm{TiO}_2\), the stoichiometric calculations reveal how to scale up the input of raw materials.
Such precise calculations prevent excess waste, ensuring resources are efficiently used without compromising the production targets.
Material Balance
Material balance or mass balance is a key principle in process engineering. It is the concept of balancing inputs and outputs in chemical reactions and unit operations. The main idea is what enters a process must either come out or accumulate within it.
In this exercise, we address material balance by calculating both the ore and the sulfuric acid required to achieve the desired output of 1000 kg of \(\mathrm{TiO}_2\).
Using principles of conservation, we estimate that 4625 kg of ore and 8661 kg of sulfuric acid are necessary to meet this output. By assuming a 50% excess of the acid, the calculations ensure all reactants fully participate in forming products, illustrating sound material management in chemical processes.

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Most popular questions from this chapter

Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right),\) an important solvent and intermediate in the production of many other chemicals, is produced by bubbling chlorine gas through liquid benzene in the presence of ferric chloride catalyst. In an undesired side reaction, the product is further chlorinated to dichlorobenzene, and in a third reaction the dichlorobenzene is chlorinated to trichlorobenzene. The feed to a chlorination reactor consists of essentially pure benzene and a technical grade of chlorine gas (98 wt\% \(\mathrm{Cl}_{2}\), the balance gaseous impurities with an average molecular weight of 25.0 ). The liquid output from the reactor contains \(65.0 \mathrm{wt} \% \mathrm{C}_{6} \mathrm{H}_{6}, 32.0 \% \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}, 2.5 \% \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) and \(0.5 \%\) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}_{3} .\) The gaseous output contains only \(\mathrm{HCl}\) and the impurities that entered with the chlorine. (a) You wish to determine (i) the percentage by which benzene is fed in excess, (ii) the fractional conversion of benzene, (iii) the fractional yield of monochlorobenzene, and (iv) the mass ratio of the gas feed to the liquid feed. Without doing any calculations, prove that you have enough information about the process to determine these quantities. (b) Perform the calculations. (c) Why would benzene be fed in excess and the fractional conversion kept low? (d) What might be done with the gaseous effluent? (e) It is possible to use 99.9\% pure ("reagent-grade") chlorine instead of the technical grade actually used in the process. Why is this probably not done? Under what conditions might extremely pure reactants be called for in a commercial process? (Hint: Think about possible problems associated with the impurities in technical grade chemicals.)

A catalytic reactor is used to produce formaldehyde from methanol in the reaction $$\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{HCHO}+\mathrm{H}_{2}$$ A single-pass conversion of \(60.0 \%\) is achieved in the reactor. The methanol in the reactor product is separated from the formaldehyde and hydrogen in a multiple-unit process. The production rate of formaldehyde is 900.0 kg/h. (a) Calculate the required feed rate of methanol to the process ( \(\mathrm{kmol} / \mathrm{h}\) ) if there is no recycle. (b) Suppose the unreacted methanol is recovered and recycled to the reactor and the single-pass conversion remains 60\%. Without doing any calculations, prove that you have enough information to determine the required fresh feed rate of methanol (kmol/h) and the rates (kmol/h) at which methanol enters and leaves the reactor. Then perform the calculations. (c) The single-pass conversion in the reactor, \(X_{\mathrm{sp}},\) affects the costs of the reactor \(\left(C_{\mathrm{r}}\right)\) and the separation process and recycle line \(\left(C_{\mathrm{s}}\right) .\) What effect would you expect an increased \(X_{\mathrm{sp}}\) would have on each of these costs for a fixed formaldehyde production rate? (Hint: To get a \(100 \%\) singlepass conversion you would need an infinitely large reactor, and lowering the single-pass conversion leads to a need to process greater amounts of fluid through both process units and the recycle line.) What would you expect a plot of \(\left(C_{\mathrm{r}}+C_{\mathrm{s}}\right)\) versus \(X_{\mathrm{sp}}\) to look like? What does the design specification \(X_{\mathrm{sp}}=60 \%\) probably represent?

Draw and label the given streams and derive expressions for the indicated quantities in terms of labeled variables. The solution of Part (a) is given as an illustration. (a) A continuous stream contains 40.0 mole\% benzene and the balance toluene. Write expressions for the molar and mass flow rates of benzene, \(\dot{n}_{\mathrm{B}}\left(\operatorname{mol} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right)\) and \(\dot{m}_{\mathrm{B}}\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right),\) in terms of the total molar flow rate of the stream, \(\dot{n}(\mathrm{mol} / \mathrm{s})\) (b) The feed to a batch process contains equimolar quantities of nitrogen and methane. Write an expression for the kilograms of nitrogen in terms of the total moles \(n(\) mol) of this mixture. (c) A stream containing ethane, propane, and butane has a mass flow rate of \(100.0 \mathrm{g} / \mathrm{s}\). Write an expression for the molar flow rate of ethane, \(\dot{n}_{\mathrm{E}}\left(\text { Ib-mole } \mathrm{C}_{2} \mathrm{H}_{6} / \mathrm{h}\right)\), in terms of the mass fraction of this species, \(x_{\mathrm{E}}\). (d) A continuous stream of humid air contains water vapor and dry air, the latter containing approximately 21 mole \(\% \mathrm{O}_{2}\) and \(79 \% \mathrm{N}_{2}\). Write expressions for the molar flow rate of \(\mathrm{O}_{2}\) and for the mole fractions of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2}\) in the gas in terms of \(\dot{n}_{1}\left(\mathrm{lb}-\mathrm{mole} \mathrm{H}_{2} \mathrm{O} / \mathrm{s}\right)\) and \(\dot{n}_{2}(\text { lb- mole dry air/s })\) (e) The product from a batch reactor contains \(\mathrm{NO}, \mathrm{NO}_{2},\) and \(\mathrm{N}_{2} \mathrm{O}_{4} .\) The mole fraction of \(\mathrm{NO}\) is 0.400. Write an expression for the gram-moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) in terms of \(n(\mathrm{mol}\) mixture) and \(y_{\mathrm{NO}_{2}}\left(\operatorname{mol} \mathrm{NO}_{2} / \mathrm{mol}\right)\)

Ethanol can be produced commercially by the hydration of ethylene: $$\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$$ Some of the product is converted to diethyl ether in the side reaction $$2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}$$ The feed to the reactor contains ethylene, steam, and an inert gas. A sample of the reactor effluent gas is analyzed and found to contain 43.3 mole\% ethylene, 2.5\% ethanol, 0.14\% ether, 9.3\% inerts, and the balance water. (a) Take as a basis 100 mol of effluent gas, draw and label a flowchart, and do a degree-of-freedom analysis based on atomic species to prove that the system has zero degrees of freedom. (b) Calculate the molar composition of the reactor feed, the percentage conversion of ethylene, the fractional yield of ethanol, and the selectivity of ethanol production relative to ether production. (c) The percentage conversion of ethylene you calculated should be very low. Why do you think the reactor would be designed to consume so little of the reactant? (Hint: If the reaction mixture remained in the reactor long enough to use up most of the ethylene, what would the main product constituent probably be?) What additional processing steps are likely to take place downstream from the reactor?

A fuel oil is fed to a furnace and burned with \(25 \%\) excess air. The oil contains \(87.0 \mathrm{wt} \% \mathrm{C}, 10.0 \% \mathrm{H},\) and 3.0\% S. Analysis of the furnace exhaust gas shows only \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}, \mathrm{SO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\). The sulfur dioxide emission rate is to be controlled by passing the exhaust gas through a scrubber, in which most of the \(\mathrm{SO}_{2}\) is absorbed in an alkaline solution. The gases leaving the scrubber (all of the \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{CO}_{2}\), and some of the \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{SO}_{2}\) entering the unit) pass out to a stack. The scrubber has a limited capacity, however, so that a fraction of the furnace exhaust gas must be bypassed directly to the stack. At one point during the operation of the process, the scrubber removes \(90 \%\) of the \(\mathrm{SO}_{2}\) in the gas fed to it, and the combined stack gas contains 612.5 ppm (parts per million) \(\mathrm{SO}_{2}\) on a dry basis; that is, every million moles of dry stack gas contains 612.5 moles of \(\mathrm{SO}_{2}\). Calculate the fraction of the exhaust bypassing the scrubber at this moment.
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