Question

Titanium ore contains Titanium(IV) Oxide, TiO2. During the production of titanium metal, this compound is first converted to Titanium(IV) chloride:

TiO2(s) + C(s) + 2 Cl2(g) = TiCl4(g) + CO2(g)

a) Identify the limiting reagent if 40g of Titanium(IV) oxide, 7g of carbon and 30g of chlorine mix.
b) What mass of Titanium(IV) chloride can be produced from these quantities?

Answer 1

You'll want to consult a periodic table to find all the molar masses and convert grams of substance to moles of substance.

Answer 2

TiO2 + C + 2Cl2 = TiCl4 + CO2
1 mol 1 mol 2mol 1 mol 1 mol
79.87g/mol 12.01g/mol 70.9g/mol 189.67g/mol 44.01g/mol
40g 7g 30g ?
Converting grams into moles

nTiO2= 40g X 1mol/79.87g = 0.500moles

nCarbon= 7.0g X 1mol/12.01g= 0.5828moles

nCl2= 30g X 1mol/70.9g= 0.423moles

stoichiometric amounts of the following substances.

nTiO2 = 0.5828mol X 1mol TiO2/1 mol C = 0.5828moles

nCarbon= 0.423mol X 1mol C/ 2mol Cl2 = 0.2115 moles

nCl2= 0.500mol X 2 mol Cl2/1mol TiO2 = 0.25 moles.

whats my limiting reagent if my moles are right

mTiCl2 = 0.25 X 189.67 = 23.70g

the correct answer is 40.1g


HELP PLZZZZZ!!!!!

Answer 3

look at your mole ratios. You need a ratio of 1:1:2 for TiO2 : C : Cl2, but you have .5 : .58 : .4. You need twice as much chlorine as titanium oxide, but you have less than a 1:1 ratio there.

Answer 4

to be honest i don't even know i did did my moles calculations right and thats why i'm not getting the right answer...

Answer 5

The number of moles of Ti-chloride will be half the number of moles of chlorine gas according to the stoichiometry.

Answer 6

Your calculations look right. You have about .4 moles of chlorine gas, so that will yield about .2 moles of Ti-chloride product.

Answer 7

and what about the limiting reagent????

Answer 8

That's the chlorine gas. Like I said, you need a 1:2 ratio of TiO2 : Cl2, but you have less than 1:1. You would need more than 60g of Cl2 for a complete reaction.
p.s. double-checking your calculations, there is an error in the last line: You have mTiCl2 = .25 * 189 = 23.7g, which is way off. It is TiCl4, and the ratio is .21 * 189.7 = about 40-something grams of TiCl4.

Answer 9

it says the limiting reagent should be chlorine... but i have two limiting reagents one is chlorine and the second is carbon because both are more in quantity than actually need to make a complete reaction....

Answer 10

The carbon is fine, it is in excess. The chlorine is less than half of what is needed.

Answer 11

ok so how do i get the mass of the TiCl4???plz