Question

Equal weights of mercury and iodine are allowed to react completely to form a mixture of mercurous and mercuric iodine leaving none of the reactants. The ratio by mole of $H{g_2}{I_2}$and $Hg{I_2}$ formed (Hg=200, I=127) is:
(A) 1: 2.7
(B) 2.7:
(C) 1: 3
(D) 3: 1

Answer 1

The basics that we need to solve this question is the formula to calculate the number of moles where the number of moles is given weight in grams by atomic weight of atom or molecular weight of molecule respectively.

Complete step by step solution:
The basics that we need to solve this question is the formula to calculate the number of moles:
$Number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}atom/molecule{\text{ }} = \;\dfrac{{wt(gram)}}{{atomic.wt/molecular.wt}}$
The number of moles of atom or molecule is given by weight in grams by the atomic weight of atom or molecular weight of molecule respectively.
-Let us assume $H{g_2}{I_2}$ to be of 1 mole.
We should use the formula above, the weight must be 754 in order to produce 1 mole of $H{g_2}{I_2}$ as the molecular weight of $H{g_2}{I_2}$ is 754.
Let's see how molecular weight is 754:
$Molecular{\text{ }}weight{\text{ }} = {\text{ }}\left[ {atomic{\text{ }}weight{\text{ }}of{\text{ }}Hg{\text{ }} + {\text{ }}atomic{\text{ }}weight{\text{ }}of{\text{ }}I} \right]$
$= 400 + 254$
So, we can come to a conclusion that, to produce 1 mole of $H{g_2}{I_2}$ it needs 400g of Hg and 254g of I.
-let number of mole of $Hg{I_2}$ be X
Let substitute the value, here number of moles is X and molecular weight is 454 as
$Molecular{\text{ }}weight{\text{ }} = {\text{ }}\left[ {atomic{\text{ }}weight{\text{ }}of{\text{ }}Hg{\text{ }} + {\text{ }}atomic{\text{ }}weight{\text{ }}of{\text{ }}I} \right]$
$= 200 + 254$
Now, $Weight{\text{ }}in{\text{ }}gram{\text{ }} = {\text{ }}number{\text{ }}of{\text{ }}moles \times {\text{ }}molecular{\text{ }}weight$
$= X(200 + 254)$
$= 200X + 254X$
In order to produce X moles of $Hg{I_2}$ it needs 200Xg of Hg and 254Xg of I.
So, Total mass of mercury required = $400 + 200X$
Total mass of iodine required = $254 + 254X$
Here according to question, the mass of mercury required is equal to mass of iodine
$$$$$$400 + 200X = 254 + 254X$$54X{\text{ }} = {\text{ }}146$$X{\text{ }} = 146 \div 54$$X = 2.7$$
The number mole of $Hg{I_2}$ is 2.7
Thus the ratio of $H{g_2}{I_2}$ and $Hg{I_2}$ formed: 1: 2.7

Hence option (A) is the correct answer.

Note: Always we should be careful while choosing the option. In question, they asked the ratio by mole of $H{g_2}{I_2}$and $Hg{I_2}$ formed so the answer is 1: 2.7. If they would have asked the ratio by mole of $Hg{I_2}$ and $H{g_2}{I_2}$ then the answer would be 2.7:1.