# What mass of sodium metal is required to make a mass of 936*g sodium chloride?

Sep 29, 2017

Approx. $400 \cdot g$ of metal are required........

#### Explanation:

We interrogate the redox reaction....

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

We gots $\frac{0.568 \cdot k g}{70.90 \cdot g \cdot m o {l}^{-} 1} = 8.01 \cdot m o l$ with respect to dichlorine....

And we gots $\frac{936 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1} = 16.01 \cdot m o l$ with respect to sodium chloride....

And given the stoichiometry, CLEARLY, we require $16.01 \cdot m o l$ sodium metal.....

...which represents a mass of.....

$16.01 \cdot m o l \times 22.99 \cdot g \cdot m o {l}^{-} 1 \cong 400 \cdot g$