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Laws of Mass Conservation

Equal weights...

Question

Equal weights of mercury and iodine are allowed to react completely to form a mixture of mercurous and mercuric iodide leaving none of the reactants. The ratio by mole of $H g_{2} I_{2}$ and $H g I_{2}$ formed (Hg= $200$ , I= $127$ ) is :

A

1 : 2.7

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B

2.7 : 1

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C

1 : 3

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D

3 : 1

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Solution

The correct option is D $1 : 2.7$
The molecular weights of $H g$ and iodine are $200$ g/mol and $254$ g/mol respectively.
To produce $1$ mole of $H g_{2} I_{2}$ , $400$ g of Hg and $254$ g of iodine are required.
To produce $x$ mole of $H g I_{2}$ , $200 x$ g of mercury and $254 x$ g of iodine are required.
Total mass of mercury required is $200 x + 400$ .
Total mass of iodine required is $254 x + 254$ .
The mass of mercury required is equal to the mass of iodine required.
Hence, $200 x + 400 = 254 x + 254$ .
Hence, $x = 2.7$ .
The ratio of the number of moles of $H g_{2} I_{2}$ and $H g I_{2}$ produced is $1 : 2.7$ .

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