Problem: The toxic pigment called white lead, Pb3(OH)2(CO3)2, has been replaced in white paints by rutile, TiO2. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO3) by mass?2FeTiO3 + 4HCl + Cl2 ⟶ 2FeCl3 + 2TiO2 + 2H2O

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The mass of rutile (TiO2) that can be produced is:

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Problem Details

The toxic pigment called white lead, Pb3(OH)2(CO3)2, has been replaced in white paints by rutile, TiO2. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO3) by mass?

2FeTiO3 + 4HCl + Cl2 ⟶ 2FeCl3 + 2TiO2 + 2H2O

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