What mass of salt would result from the reaction between the alkali metal, and 10*mol "chlorine gas"?

Feb 20, 2018

Just under $1200 \cdot g$ of salt would result....

Explanation:

We write the stoichiometric equation...

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

And so with respect to each equiv of dichlorine gas we GETS 2 equiv of salt...and since we used $10 \cdot m o l$ $C {l}_{2}$ we get approx. $20 \cdot m o l$ salt...

Thus 20*molxx58.44*g*mol^-1=??*g...