In a certain operation 358 g of ${\mathrm{TiCl}}_4$ is reacted with 96 g of Mg. Calculate % yield of Ti if 32 g of Ti is actually obtained [At. wt. Ti = 48, Mg = 24] $\left[\mathrm{Hint}:\frac{358}{190}=1.88\right]$

  $$\begin{gathered} {\mathrm{TiCl}}_4+2\mathrm{Mg}\to \mathrm{Ti}+2{\mathrm{MgCl}}_2 \\ 1\mathrm{mole} 2\mathrm{moles} 1\mathrm{mole}\left(48\mathrm{g}\right) \\ \mathrm{given}; \frac{358}{190} \frac{96}{24} ? \\ =1.88 =4 \mathrm{moles} \\ \mathrm{So} {\mathrm{TiCl}}_4 \mathrm{is} \mathrm{limiting} \mathrm{reagent} \\ 1\mathrm{mole} \mathrm{of} {\mathrm{TiCl}}_4\to 48\mathrm{g} \mathrm{of} \mathrm{Ti} \\ 1.88\mathrm{moles} \mathrm{of} {\mathrm{TiCl}}_4\to ?\mathrm{a} \end{gathered}$$

   mass of ‘Ti ’obtained  = $1.88\times 48\mathrm{g}$

But actual mass of ‘Ti’ obtained = 32g

% yield of ‘Ti’ = $\frac{32}{48\times 1.88}\times 100=35.4\%$