Question

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

Answer 1

Approx. `41*g` of salt will be produced.

Explanation:

We interrogate the reaction:

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

One equiv of sodium metal reacts with half an equiv of `Cl_2(g)` to give one equiv of salt.

`"Moles of sodium"=(75*g)/(22.99*g*mol^-1)=3.26*mol`

`"Moles of chlorine gas"=(25*g)/(70.90*g*mol^-1)=0.353*mol`.

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

And thus AT MOST, only `2xx0.353*mol` `NaCl` can be generated, i.e. `2xx0.353*cancel(mol)xx58.44*g*cancel(mol^-1)=41.3*g`