When Sn reacts with calculated amount of HgCl2 and exce

When Sn reacts with calculated amount of $H g C l_{2}$ and excess $H g C l_{2}$ , then the product will be respectively.

$S n C l_{2}, S n C l_{4}$

$S n C l_{4}, S n C l_{2}$

$S n C l_{4} o n l y$

$S n C l_{2} o n l y$

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Right Answer is: A

SOLUTION

This is one of the cases in chemistry where you will see that the amount of reactant is very important to get the product.
When Sn is heated with calculated amount of $H g C l_{2}$ , anhydrous $S n C l_{2}$ is formed.
$S n + H g C l_{2} \to S n C l_{2} + H g$
and, by heating Sn with excess of $H g C l_{2}$ , Stannic Chloride is formed.

$S n + 2 H g C l_{2} \to S n C l_{4} + 2 H g$

When Sn reacts with calculated amount of HgCl2 and exce

Right Answer is: A

SOLUTION

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