Answer:
2928kg of ore are required.
2585kg of the 80% H₂SO₄ solution are required.
Explanation:
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:
<em>Moles TiO₂ -Molar mass: 79.866g/mol-:</em>
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) =
<h3>2928kg of ore are required</h3><h3 />
<em>Mass 80% sulfuric acid:</em>
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) =
<h3>2585kg of the 80% H₂SO₄ solution are required</h3>
