Question

0.858 g of TiCl4 reacts with excess oxygen according to the equation
TiCl4 (s) + O2 (g)  TiO2 (s) + 2Cl2 (g)
What is the percentage yield of TiO2 if 0.287 g of TiO2 was formed?
Molar masses of TiCl4 and TiO2 are, respectively, 189.67 and 79.87 g/mol

Answer 1

$\begin{aligned} TiCl_{4_(s)} + O_{2_(g)} \longrightarrow TiO_{2_(s)} + 2Cl_{2_(g)} \end{aligned}$

According to the chemical reaction above;

1 mole of $TiCl_4$ produces 1mole of $TiO_2$

0.857g of $TiCl_4$ = $\frac{0.858}{189.67}mol = 0.0452mol$

$\therefore$ 0.00452 mol of $TiCl_4$ should produce 0.00452 mol of $TiO_2$

​

0.00452 mol of $TiO_2$ = 0.0452(79.87)g = 0.361g

Therefore 0.361g of $TiO_2$ is the theoretical yield;

and 0.287g is the amount of $TiO_2$ actually produced

$\begin{aligned} \% yield = \dfrac{Actual\ yield}{Theoretical\ yield} × 100\% \end{aligned}$ $\begin{aligned} \\ \% yield = \dfrac{0.287}{0.361} × 100\% \end{aligned}$

$$

$\begin{aligned} \% yield = 79.5 \% \end{aligned}$