Asked by Anonymous
If 0.790 mol of solid TiO2 and 8.53 g of solid C are reacted stoichiometrically according to the balanced equation, how many moles of solid TiO2 remain?
3TiO2(s) + 4C(s) + 6Cl2(g) → 3TiCl4(l) + 2CO2(g) + 2CO(g)
3TiO2(s) + 4C(s) + 6Cl2(g) → 3TiCl4(l) + 2CO2(g) + 2CO(g)
Answers
Answered by
DrBob222
Change C to moles.
8.53 g/12.01 = ?? mols
Now convert moles TiO2 to moles of any one of the products, say TiCl4.
Convert moles C to moles TiCl4.
That will give you the limiting reagent; either TiO2 or C.
Using the limiting reagent (which I assume will be C), convert moles C to moles TiO2 and subtract from the initial moles of TiO2 to find the amount of TiO2 remaining after reaction.
8.53 g/12.01 = ?? mols
Now convert moles TiO2 to moles of any one of the products, say TiCl4.
Convert moles C to moles TiCl4.
That will give you the limiting reagent; either TiO2 or C.
Using the limiting reagent (which I assume will be C), convert moles C to moles TiO2 and subtract from the initial moles of TiO2 to find the amount of TiO2 remaining after reaction.
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