What mass of salt can be isolated from a 75*g mass of sodium metal, and a 25*g mass of chlorine gas?

Mar 22, 2017

Approx. $41 \cdot g$ of salt will be produced.

Explanation:

We interrogate the reaction:

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

One equiv of sodium metal reacts with half an equiv of $C {l}_{2} \left(g\right)$ to give one equiv of salt.

$\text{Moles of sodium} = \frac{75 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 3.26 \cdot m o l$

$\text{Moles of chlorine gas} = \frac{25 \cdot g}{70.90 \cdot g \cdot m o {l}^{-} 1} = 0.353 \cdot m o l$.

So sodium metal is present in wast stoichiometric excess, and chlorine gas is the limiting reagent.

And thus AT MOST, only $2 \times 0.353 \cdot m o l$ $N a C l$ can be generated, i.e. $2 \times 0.353 \cdot \cancel{m o l} \times 58.44 \cdot g \cdot \cancel{m o {l}^{-} 1} = 41.3 \cdot g$