Question

In a certain reaction 389.5 g of $TiC{l}_4$ is reacted with 96 g of $Mg$. Calculate the percentage yield of $Ti$ if 32 g of $Ti$ is actually obtained . $[At.wtTi=48u,Mg=24uCl=35.5u]$

Answer 1

The corresponding reaction is
$TiC{l}_4+2Mg\to Ti+2MgC{l}_2$

Molar mass of $TiC{l}_4=48+4\times 35.5=190g$
Mass of $TiC{l}_4$ given is 389.5 g
Moles of $TiC{l}_4=\frac{389.5}{190}=2.05$ mol
Given, mass of $Mg=96g$
$\therefore$ Moles of $Mg=\frac{96}{24}$ mol $=4$ mol
From the reaction,
2 moles of Mg reacts with 1 mole of $TiC{l}_4$
Thus 4 mol of Mg will react with 2 moles $TiC{l}_4$
So $Mg$ is the limiting reagent here.
So the moles of $Ti$ produced = 2 mol
Hence, theoretical mass of $Ti$ produced = $96g$
obtained mass of Ti = 32 g.
$\therefore$ yield $=\frac{32}{96}\times 100=33.33\%$