Titanium dioxide (TiO2) is used extensively as a white pigment. It is produced from an ore that contains ilmenite (FeTiO3) and ferric oxide (Fe2O3). The ore is digested with an aqueous sulfuric acid solution to produce an aqueous solution of titanyl sulfate [(TiO)SO4 ] and ferrous sulfate (FeSO4 ). Water is added to hydrolyze the titanyl sulfate to H2TiO3, which precipitates, and H2SO4. 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% H2SO4 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 [Fe2 (SO4 )3 ]. 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 kg of pure TiO2.

To solve this question we need first to find the moles of titanium in 1000kg of TiO₂. Keeping in mind the 89% of descomposition we can find the mass of the ore and the mass of the 80% sulfuric acid required:

Moles TiO₂ -Molar mass: 79.866g/mol-:

1x10⁶g * (1mol / 79.866g) = 12521 moles Titanium

In mass -Molar mass Ti: 47.867g/mol-:

12521 moles Titanium * (47.867g / mol) = 599341.4g of Ti.

As the ore contains 24.3% of Ti:

599341.4g of Ti = 599.34kg Ti * (100 / 24.3) = 2606kg ore

As the descomposition is just of 89%:

2606kg ore * (100 / 89) =

2928kg of ore are required

Mass 80% sulfuric acid:

12521 moles Titanium = 12521 moles H₂SO₄ * (100/89) = 14068.5 moles of H₂SO₄ are required.

In an excess of 50% =

14068.5 moles of H₂SO₄ are required * 1.5 = 21102.8 moles of H₂SO₄.

The mass is:

21102.8 moles of H₂SO₄ * (98g / mol) = 2068075g = 2068kg of sulfuric acid

That is in the 80%:

2068kg of sulfuric acid * (100/ 80) =

2585kg of the 80% H₂SO₄ solution are required

mickymike92 mickymike92 · Answer 2 · 2022-01-04T15:53:29+01:00

2771.53 kg of titanium ore and 2584.63 kg of 80% H₂SO₄ are required to produce 1000 kg of pure TiO2.

The moles of titanium in 1000 kg of TiO₂ is first determined:

Molar mass of TiO₂ = 79.88 g/mol

Molar mass of Ti = 47.88 g/mol

moles of TiO₂ = mass of TiO₂/molar mass of TiO₂

mass of TiO₂ = 1000 kg = 1 * 10⁶g

moles of TiO₂ = 1 * 10⁶g/79.88 g/mol

moles of TiO₂ = 12518.77 moles

1 mole of TiO₂ produces 1 mole of titanium

12518.77 moles TiO₂ produces 12518.77 moles of titanium

mass of titanium = number of moles * molar mass

mass of titanium = 12518.77 moles * 47.88 g/mol

mass of titanium = 599398.70 g

Since the percentage mass of titanium in the ore is 24.3%

mass of ore = 599399.18 g * 100/24.3

mass of ore = 2466661.31 g

mass of ore in kg = 2466.66 kg

Since only 89% of ore decomposed, mass of ore required = 2466.66 kg *100/89

mass of ore required = 2771.52

Mass 80% sulfuric acid required is calculated as follows:

Molar mass of H₂SO₄ = 98.00 g/mol

1 mole of titanium requires 1 mole of H₂SO₄

Therefore, 12521 moles Titanium = 12521 moles H₂SO₄

Since only 89% of the ore decomposes;

moles of H₂SO₄ required = 12518.77 moles * 100/89

moles of H₂SO₄ required = 14066.03 moles

The H₂SO₄ solution is supplied in excess of 50%

moles of H₂SO₄ supplied = 14066.03 moles * 150/100

moles of H₂SO₄ supplied = 21099. 04 moles

mass of H₂SO₄ supplied = number of moles * molar mass

mass of H₂SO₄ supplied = 21099. 04 moles * 98 g/mol

mass of H₂SO₄ supplied = 2067702 g

Since the solution is only 80% H₂SO₄;

mass of 80% H₂SO₄ required = 2067702 g * 100/80

mass of 80% H₂SO₄ required = 2584627.5 g

mass of 80% H₂SO₄ required in kg = 2584.63 kg

Therefore, 2771.53 kg of ore and 2584.63 kg of 80% H₂SO₄ are required to produce 1000 kg of pure TiO2.

Learn more about purification of ores and sulfuric acid at: https://brainly.com/question/947373