# What is the maximum mass of "TiCl"_4 that can be obtained from 60.5 g "TiO"_2, 59.9 g "Cl"_2 and excess carbon?

## Give your answer in grams to three significant figures. The balanced chemical equation for the reaction is given below. $3 \text{TiO"_2(s)+6"Cl"_2+4"C"(s)->3"TiCl"_4(g)+2"CO"_2(g)+2"CO} \left(g\right)$

Oct 1, 2017

$80.2 g$

#### Explanation:

With excess carbon we only need to determine whether the amount of chlorine is limiting or not. Then we calculate the produce $T i C {l}_{4}$ from the known moles of chlorine.
We need two moles of $C {l}_{2}$ for every $T i$ atom.
$\frac{59.9}{70.9} = 0.845$ moles $C {l}_{2}$
(we could also do the calculation precisely using a Cl atom and 35.45g/mol, but we already know it is doubled in this reaction).

$\frac{60.5}{47.87} = 1.264$ moles of Ti. Therefore, $C {l}_{2}$ is limiting.

We can ONLY produce $\frac{0.845}{2} = 0.423$ moles of $T i C {l}_{4}$ formed from the same number of moles of $T i {O}_{2}$.

The mass is $0.423 m o l \times 189.67 \left(\frac{g}{m o l}\right) = 80.23 g$ or just $80.2 g$ for three significant figures.