# Given a 3.00*g mass of sodium metal, what mass of chlorine gas is equivalent? What mass of sodium chloride could be made given this quantity?

Sep 11, 2017

We assume a stoichiometric excess of chlorine gas......and we work out the mass of sodium chloride produced given 100% yield.

#### Explanation:

....and we must write a stoichiometric equation....

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

As written, this is a redox reaction, and sodium metal is oxidized by half an equiv of (binculear) chlorine gas....

$\text{Moles of natrium} = \frac{3.00 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 0.131 \cdot m o l$

And thus we should get $0.131 \cdot m o l$ of salt.....i.e.

$0.131 \cdot m o l \times 58.44 \cdot g \cdot m o {l}^{-} 1 = 7.63 \cdot g$.