Question

In a certain operation 358 gm of $TiC{l}_4$ is reacted with 96 gm of $Mg$. The $\%$ yield of $Ti$ if 32 gm of $Ti$ actually obtained is :

[Atomic weight $Ti=48,Mg=24$]

• A. $35.38\%$
• B. $66.5\%$
• C. $100\%$
• D. $60\%$

Answer 1

The correct option is A $35.38\%$

$TiC{l}_4+2Mg\to Ti+2MgC{l}_2$
Initial mole of $TiC{l}_4=\frac{358}{190}=1.88$ and of $Mg=\frac{96}{24}=4$
Final mole of $TiC{l}_4=0$ and that of $Mg=4-(2\times 1.88)$
Final mole of $Ti=1.88$ and that of $Mg=2\times 1.88$
Weight of Ti obtained $=\frac{358}{190}\times 48$
Hence, the $\%$ yield $=\frac{32\times 100}{\frac{358\times 48}{190}}=35.38\%$