Question

A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium

Answer 1

A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium

Explanation:

Ore contains ---- 65.2% $TiO_2$

Mass% of titanium in TiO2 can be calculated as shown below:

$mass percentage of Ti in TiO2=\frac{mass of Ti}{mass of TiO_2} *100\\=(47.86g/79.866g)* 100\\=59.9$

Given 10.0 metric tons of titanium is required.

The mass of ore that should be processed can be calculated as shown below:

Mass of Ti = ore x TiO2 % x Ti mass %

10.0 x 1000 kg = M (mass of ore) x (65.2/100) x (59.9/100) Ti

=>M=(10.0 metric tons) / (0.652 x 0.599)

=>M=25.6 metric tons

Hence, the mass of ore required is 25.6 metric tons.