Question

The toxic pigment called whte lead Pb3(OH)2(CO33)2 has been replaced in white paints by rutile, TiO2. How much rutile can be prepared from 379 g of an ore that contains 88.3% ilemnite (FeTiO3) by mass?

2FTiO3 + 4HCl + Cl2 ? 2FeCl3 + 2TiO2 + 2H2O

Answer 1

Answer : The mass of
`TiO_2` prepared can be 176.48 grams.

Explanation :

First we have to calculate the mass of
`FeTiO_3`.

As we are given that the 88.3 %
`FeTiO_3` by mass. That means, 88.3 grams of
`FeTiO_3` present in 100 grams of solution.

`\text{Mass of }FeTiO_3=(88.3)/(100)* 379=334.657g`

Now we have to calculate the moles of
`FeTiO_3`.

Molar mass of
`FeTiO_3` = 151.7 g/mole

`\text{Moles of }FeTiO_3=\frac{\text{Mass of }FeTiO_3}{\text{Molar mass of }FeTiO_3}=(334.657g)/(151.7g/mole)=2.206moles`

Now we have to calculate the moles of
`TiO_2`.

The balanced chemical reaction is,

`2FeTiO_3+4HCl+Cl+2\rightarrow 2FeCl_3+2TiO_2+2H_2O`

From the balanced reaction, we conclude that

As, 2 moles of
`FeTiO_3` react to give 2 moles of
`TiO_2`

So, 2.206 moles of
`FeTiO_3` react to give 2.206 moles of
`TiO_2`

Now we have to calculate the mass of
`TiO_2`.

`\text{Mass of }TiO_2=\text{Moles of }TiO_2* \text{Molar mass of }TiO_2`

Molar mass of
`TiO_2` = 80 g/mole

`\text{Mass of }TiO_2=(2.206mole)* (80g/mole)=176.48g`

Therefore, the mass of
`TiO_2` prepared can be 176.48 grams.