Question

Equal weights of mercury and $l_2$ are allowed to react completely to form a mixture of mercurous and mercuric iodide leaving none of the reactants. Calculate the ratio of the weights of $H{g}_2{l}_2$ and $Hg{l}_2$ formed.

• A. $1:0653$
• B. $0.732:1$
• C. $1:0.523$
• D. $0.532:1$

Answer 1

The correct option is D $0.532:1$

Let the weight of both $Hg$ & $I_2$ be $x$

Now we have $127$ moles of $Hg$ & $200$ moles of $I$. Say out of $127$ moles, $Y$ moles of $Hg$ gets converted into $H{g}_2{I}_2$.

They $Y$ moles of $I$ will also be consumed.

Now, $Hg$ left= $127-Y$ moles

$I$ left= $200-Y$ moles

Now to form $Hg{I}_2$ the moles of $I$ left should be double than that of $Hg$ for complete conversion.

$\therefore 200-Y=2(127-Y)$

$\Rightarrow Y=54$

This shows that $54$ moles of $Hg$ will form $Hg{I}_2$ meaning $27$ moles of $H{g}_2{I}_2$ & rest will form $Hg{I}_2\Rightarrow 73$ moles of $Hg{I}_2$ .

Now the ratio of weights,

$=\frac{27(200\times 2+2\times 127)}{73(200+2\times 127)}$

$=0.532:1$