Question #14d26

Mar 2, 2017

$2 N a {I}_{a q} + C {l}_{2} \to 2 N a C {l}_{a q} + {I}_{2}$

Explanation:

In a single replacement reaction, the negative ions ${I}^{-} 1 \mathmr{and} C {l}^{-}$ swap places.

Chlorine gas and iodine are diatomic, meaning they are found in nature as $C {l}_{2}$ and ${I}_{2}$.

Write the skeleton equation:
$N a I + C {l}_{2} \to N a C l + {I}_{2}$ (not balanced)

A balanced equation requires the number of atoms of each element to be equal on each side of the equation.

Since there are $2$ Chlorine atoms on the left side of the equation, there needs to be a $2$ placed in front of the $N a C l$ which represents $2$ molecules of sodium chloride. This then requires a $2$ to be placed in front of the $N a I$.

$2 N a {I}_{a q} + C {l}_{2} \to 2 N a C {l}_{a q} + {I}_{2}$