Answer:
b. lithium
Explanation:
Li the least likely, to lose an electron.
Be-beryllium have 2 electrons and it is in the 2 nd period
Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.
next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.
use the mols of the limiting reagent to compare with the mols of the product.
take the mols of the product/mr of the product.
this will give u the mass.
Explanation:
The relation between
is given by :

Where :
= Ionic prodcut of water
The value of the first ionization constant of sodium sulfite = 
The value of
:


The value of the second ionization constant of sodium sulfite = 
The value of
:


It would have to be 40.0 only because it wouldn’t add up