Question

What mass of salt will be obtained when a `18.0*g` mass of sodium reacts with a `23.0*g` mass of chlorine gas?

Answer 1

Approx. `38*g`............

Explanation:

We interrogate the chemical reaction.....

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

And clearly, the stoichiometry is `1:1/2`, where we deal with `Cl_2` (you simply have to know that the elemental gases, save the Noble gases, are BINUCLEAR, i.e. `X_2`)

So we take the molar quantities by dividing each mass thru by its atomic mass, which we get from the Periodic Table:

`"Moles of sodium"=(18.0*g)/(22.99*g*mol^-1)=0.783*mol.`

`"Moles of chlorine"=(23.0*g)/(70.9*g*mol^-1)=0.320*mol.`

Given these molar quantities, it is CLEAR that chlorine gas is in deficiency, and ONLY `0.640*mol` of sodium will react. Is it clear? Why so?

And thus, given complete reaction, we should get `0.640*mol` sodium chloride, which represents a mass of.......

`0.640*molxx58.44*g*mol^-1~=38*g` with respect to `NaCl`.