Question

Equal weights of zinc and iodine react together and the iodine is completely converted to $$ZnI_{2}$$. 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

Solution

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 $$ZnI_{2}$$, we have $$Zn$$ $$+{I}_{2}\rightarrow ZnI_{2}$$Initial number of moles of zinc and iodine is $$\displaystyle \dfrac{x}{65}$$ and $$\dfrac{x}{254}$$ moles respectively.Number of moles of Zn at the end of the reaction is $$(\displaystyle \frac{x}{65}-\frac{x}{254})$$ moles.$$\therefore$$ Fraction of Zn remained unreacted = $$\dfrac{\left (\dfrac{x}{65}-\dfrac{x}{254} \right )}{\dfrac{x}{65}}=0.74$$.Chemistry

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