Equal weights of Zn metal and iodine are mixed together and $l_{2}$ is completely converted to$Znl_{2}$. What fraction by weight of original Zn remains unreacted? (Zn = 65, I = 127)

Question

Dr. Abubacker · Accepted Answer

(b)
By the equation
$Zn + I_{2} \longrightarrow Znl_{2}$
Initial moles (if $x$ be the wt. $\frac{x}{65}\ \ \ \ \ \ \frac{x}{254}\ \ \ \ \ \ \ \ \ \ \ \ 0$
Of Zn and $I_{2}$ each initially)
No. of moles at the end $\left( \frac{x}{65} - \frac{x}{254} \right)\ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \frac{x}{254}$
Of reaction
So, fraction of $Zn$ unreacted $= \frac{\frac{x}{65} - \frac{x}{254}}{\frac{x}{65}} = 0.74$

Q. Equal weights of Zn metal and iodine are mixed together and $l_{2}$ is completely converted to$Znl_{2}$. What fraction by weight of original Zn remains unreacted? (Zn = 65, I = 127)

Explanation

Reviewed by Dr. Abubacker

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