Question

Question: Titanium dioxide,$\left(Ti{O}_2\right)$ , reacts with carbon and chlorine to give gaseous $\left(TiC{l}_4\right)$:

$Ti{O}_2+2C+2C{l}_2\to TiC{l}_4+2CO$

The reaction of 7.39 Kg titanium dioxide with excess C and chlorine gas gives 14.24g of titanium tetrachloride. Calculate the theoretical yield and its % yield.

Short answer

We are given the yield of titanium chloride which is produced when titanium dioxide reacts with excess C and chlorine gas. We need to calculate the theoretical yield and % yield involved.

Step-by-step solution

Step (1)- Process of calculation of theoretical yield

Titanium dioxide reacts with carbon and chloride to form gaseous titanium tetrachloride. The equation for this reaction is

$Ti{O}_2+2C+2C{l}_2\to TiC{l}_4+2CO$

Titanium dioxide reacts with an excess of carbon and chloride; so the limiting reactants are titanium dioxide and the theoretical yield is calculated using its moles.

Step (2): Calculation of theoretical and % yield

According to the balanced reaction, 1 mole of $TiC{l}_4$. The mass of $TiC{l}_4$ produced from 7.39kg $Ti{O}_2$ is:

$$\begin{gathered} \left(7.39\times {10}^3.Ti{O}_2\right)\times \frac{1mol.Ti{O}_2}{77.87g.Ti{O}_2}\times \frac{1moleTiC{l}_4}{1moleTi{O}_2}\times \frac{184.62g.TiC{l}_4}{1mol.TiC{l}_4} \\ =1.75\times {10}^{4}g.TiC{l}_4 \\ =17.5Kg.TiC{l}_4 \end{gathered}$$

Therefore, the theoretical yield of titanium chloride is $1.75\times {10}^{4}g$

The ratio of the actual yield to the theoretical yield multiplied by 100% gives the % yield for production in the reaction.

It is given that the actual yield of titanium tetrachloride is 14.24kg. So, the % yield of titanium tetrachloride will be

$$\begin{gathered} =\frac{14.24Kg}{17.5Kg}\times 100\% \\ =81.4\% \end{gathered}$$