# What mass of salt will be obtained when a 18.0*g mass of sodium reacts with a 23.0*g mass of chlorine gas?

Aug 21, 2017

Approx. $38 \cdot g$............

#### Explanation:

We interrogate the chemical reaction.....

$N a \left(s\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow N a C l \left(s\right)$

And clearly, the stoichiometry is $1 : \frac{1}{2}$, where we deal with $C {l}_{2}$ (you simply have to know that the elemental gases, save the Noble gases, are BINUCLEAR, i.e. ${X}_{2}$)

So we take the molar quantities by dividing each mass thru by its atomic mass, which we get from the Periodic Table:

$\text{Moles of sodium} = \frac{18.0 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 0.783 \cdot m o l .$

$\text{Moles of chlorine} = \frac{23.0 \cdot g}{70.9 \cdot g \cdot m o {l}^{-} 1} = 0.320 \cdot m o l .$

Given these molar quantities, it is CLEAR that chlorine gas is in deficiency, and ONLY $0.640 \cdot m o l$ of sodium will react. Is it clear? Why so?

And thus, given complete reaction, we should get $0.640 \cdot m o l$ sodium chloride, which represents a mass of.......

$0.640 \cdot m o l \times 58.44 \cdot g \cdot m o {l}^{-} 1 \cong 38 \cdot g$ with respect to $N a C l$.