About half the world's production of pigments for paints involves the formation of white TiO2. It is made on a large scale by the chloride process , starting with ores containing only small amounts of rutile, TiO2. The ore is treated with chlorine and carbon. this produces TiCl4 and gaeseous product according to the following equation:
3TiO2(s)+ 4C(s)+ 6Cl2(g)--> 3TiCl4(l)+ 2CO2(g)+ 2CO(g)
the titanium(IV)chloride is then converted into titanium(IV)oxide of high purity
TiCl4(l)+ O2(g)--> TiO2(s)+ 2Cl2(g)
suppose the first process can be carried out with 70.0% yield and then second one with 93.0% yield. How many kilograms of TiO2 could be produced starting with 1.00metric ton (1.00x10^6g)of an ore that is 0.75% rutile TiO2?
Here is a worked example of a stoichiometry problem Your problem has two such problems in it (or you can combine the two if you wish).
http://www.jiskha.com/science/chemistry/stoichiometry.html
Convert 1 metric ton to grams ore then to grams TiO2 (using the 0.75% in the ore), then to g TiCl4 (using the 70%%) and finally to TiO2 final product (using the 93.0%). Post your work if you get stuck.
i converted the 1metric ton to grams but then how to i convert it to grams of TiO2(using the 0.75% in the ore)
To determine the amount of TiO2 that could be produced starting with 1.00 metric ton (1.00 x 10^6g) of ore, we need to follow these steps:
1. Calculate the mass of rutile TiO2 in the ore:
Mass of ore = 1.00 metric ton = 1.00 x 10^6 g
Percentage of rutile TiO2 in the ore = 0.75%
Mass of rutile TiO2 = (Percentage of rutile TiO2 / 100) x Mass of ore
= (0.75 / 100) x 1.00 x 10^6 g
2. Calculate the molar mass of TiO2:
Molar mass of TiO2 = Atomic mass of Ti + 2 x Atomic mass of O
= 47.87 g/mol + 2 x 16.00 g/mol
= 79.87 g/mol
3. Calculate the number of moles of rutile TiO2:
Number of moles = Mass / Molar mass
= Mass of rutile TiO2 / Molar mass
4. Determine the theoretical yield of TiO2 from the first process:
The first process has a 70.0% yield, so multiply the number of moles of rutile TiO2 by 70.0/100 to get the moles of TiO2 produced.
5. Convert the moles of TiO2 to grams:
Mass of TiO2 = Moles of TiO2 x Molar mass
6. Determine the yield of TiO2 from the second process:
The second process has a 93.0% yield, so multiply the mass of TiO2 by 93.0/100 to get the final mass of TiO2 produced.
Now, let's calculate the mass of TiO2 that could be produced:
Mass of rutile TiO2 = (0.75 / 100) x 1.00 x 10^6 g = 7500 g
Number of moles of rutile TiO2 = Mass of rutile TiO2 / Molar mass
= 7500 g / 79.87 g/mol ≈ 93.84 mol
Theoretical yield of TiO2 from the first process = Number of moles of rutile TiO2 x 70.0/100
≈ 93.84 mol x 0.70 ≈ 65.69 mol
Mass of TiO2 = Moles of TiO2 x Molar mass
= 65.69 mol x 79.87 g/mol ≈ 5249 g
Final yield of TiO2 from the second process = Mass of TiO2 x 93.0/100
≈ 5249 g x 0.93 ≈ 4879.57 g ≈ 4.88 kg
Therefore, starting with 1.00 metric ton of ore that is 0.75% rutile TiO2, approximately 4.88 kg of TiO2 could be produced.