# An sodium metal salt is prepared from 3.0*g metal and 2.0*g of iodine. What is the MAXIMUM possible mass of NaI that can be prepared?

Oct 26, 2016

Well, first you need a stoichiometric equation......

#### Explanation:

$N a + \frac{1}{2} {I}_{2} \rightarrow N a I$

$\text{Moles of metal}$ $=$ $\frac{3.0 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.131 \cdot m o l$.

$\text{Moles of halogen}$ $=$ $\frac{2.0 \cdot g}{2 \times 126.9 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.24 \times {10}^{-} 4 \cdot m o l$.

Now clearly, the metal is present in vast stoichiometric excess. At most, we can form $2.48 \times {10}^{-} 4 \cdot m o l$ of salt. I leave it to you to determine the mass of salt possible.

Note that the halogens, as elements, are all binuclear, ${X}_{2}$, and this is knowledge that the examiner expects. All elemental gases, save for the Noble Gases, and (maybe) mercury, are binuclear.