CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
0.74
loader
C
0.47
loader
D
0.17
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image