Question

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

Answer 1

Just under `1200*g` of salt would result....

Explanation:

We write the stoichiometric equation...

`Na(s)+1/2Cl_2(g) rarr NaCl(s)`

And so with respect to each equiv of dichlorine gas we GETS 2 equiv of salt...and since we used `10*mol` `Cl_2` we get approx. `20*mol` salt...

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