Question

Titanium, which is used to make air plane engines and frames, can be obtained from titanium tetrachloride, which in turn is obtained from titanium oxide by the following process.
$3Ti{O}_2\left(s\right)+4C\left(s\right)+6C{l}_2\left(g\right)\to 3TiC{l}_4\left(g\right)+2C{O}_2\left(g\right)+2CO\left(g\right)$
A vessel contains $4.32$ gm $Ti{O}_2$, $5.76$ gm C and; $7.1$gm $C{l}_2$, suppose the reaction goes to completion as written, how many gram of $TiC{l}_4$ can be produced? $(Ti=48)$

Answer 1

Balanced chemical equation is

$3Ti{O}_2\left(s\right)+4C\left(s\right)+6C{l}_2\left(g\right)⟶3TiC{l}_4\left(g\right)+2C{O}_2+2CO$

Given mass of-

$Ti{O}_2=4.32g$

No. of moles=$\frac{4.32}{80}=0.054$ mol

Given mass of $C=5.76g$

No. of moles=$\frac{5.76}{12}=0.48$ mol

Given mass of $C{l}_2=\frac{7.1}{71}=0.1$ mol

Ratio of moles given and stoichiometric coefficient:

For, $Ti{O}_2$ is $\frac{0.054}{3}=0.018$

for $C$ is $\frac{0.48}{4}=0.12$

for $C{l}_2$ is $\frac{0.1}{6}=0.017$

As $0.017$ is the lowest, $C{l}_2$ is the limiting reagent and formation of product will depend on the number of moles of $C{l}_2$.

So, number of moles of $TiC{l}_4$ produced= $\frac{3}{6}\times 0.1=0.05$ moles

and, mass produced of $TiC{l}_4=0.05\times 190=9.5g$