Question

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.

a) 191.7 g
b) 214.4 g
c) 236.8 g
d) 279.5 g

Answer 1

The amount of rutile that can be prepared from 379 g of the ore is approximately 211.6 g.

Step-by-step explanation:

To determine the amount of rutile (TiO2) that can be prepared from an ore containing 88.3% ilmenite (FeTiO3), we need to calculate the mass of ilmenite in the ore and then convert it to the mass of rutile.

First, calculate the mass of ilmenite in the ore:

Mass of ilmenite = 88.3% * 379 g = 334.357 g.

Since the reaction requires 2 moles of ilmenite to produce 2 moles of rutile, the molar mass of ilmenite and rutile are respectively:

Molar mass of ilmenite = (1 * 55.845 g/mol) + (1 * 47.867 g/mol) +(3 * 16.00 g/mol) = 151.087 g/mol.

Molar mass of rutile = (1 * 47.867 g/mol) + (2 * 16.00 g/mol) = 95.734 g/mol.

Now, convert the mass of ilmenite to the mass of rutile:

Mass of rutile = (334.357 g * 95.734 g/mol) / 151.087 g/mol = 211.615 g.

Therefore, the amount of rutile that can be prepared from 379 g of the ore is approximately 211.6 g.