Question

Equal weights of zinc and iodine react together and the iodine is completely converted to $Zn{I}_2$. What fraction by weight of the original zinc remains unreacted? ($Zn=65,I=127$ g/mol)

• A. $0.6$
• B. $0.74$
• C. $0.47$
• D. $0.17$

Answer 1

The correct option is B $0.74$

Let $x$ g be the initial weight of Zn metal and iodine each.

Since $I_2$ is completely converted to $Zn{I}_2$, we have $Zn$ $+I_2\to Zn{I}_2$
Initial number of moles of zinc and iodine is $\frac{x}{65}$ and $\frac{x}{254}$ moles respectively.
Number of moles of Zn at the end of the reaction is $(\frac{x}{65}-\frac{x}{254})$ moles.
$\therefore$ Fraction of Zn remained unreacted = $\frac{(\frac{x}{65}-\frac{x}{254})}{\frac{x}{65}}=0.74$.