Question

In a certain operation $358\, g$ of $TiCl_4$ is reacted with $96$ g of $Mg$. Calculate $\%$ yield of $Ti$ if $32\, g$ of $Ti$ is actually obtained [At. wt. $Ti = 48, Mg = 24][$ Hint: $\frac{358}{190} = 1.88]$

• A. $35.38\%$
• B. $66.6\%$
• C. $100\%$
• D. $60\%$

Answer 1

Answer: (A)

$TiCl_4 + 2Mg \to Ti + 2MgCl_2$
Initial mole $\frac{358}{190} = 1.88 \,\,\frac{96}{24} = 4$
final mole $0\,\,4-2 \times 1.88 \,\,1.88 \,\,2\times 1.88$
Weight of $Ti$ obtained $ = \frac{358}{190} \times 48$
$\%$ Yield $ = \frac{\frac{32\times 100}{358 \times 48}}{190} = 35.38\%$