Reaction of acid with metal oxide. How many H+ is needed?
Oxides contain the O2- anion, in other words an oxygen atom that has gained 2 electrons. O gains these electrons from the metal when forming the oxide. Metals give away electrons. That’s one of their characteristics. This means for every O atom, the metal atoms must have given away two electrons.
Examples of metal oxides:
Na2O (sodium oxide) – Na forms a +1 ion so to balance one O2- there must be two Na atoms needed.
MgO (magnesium oxide) – One Mg can give 2e- so only one Mg was needed to supply both e- to the O2-.
Al2O3 (aluminium oxide) – Here there are three O2- atoms, so a total of 6 e- had to be supplied by the metal. Al forms a +3 ion so two Al atoms were needed to provide this, hence the formula.
MnO2 (manganese dioxide, dark brown, almost black insoluble solid) We have two O atoms, hence need a metal that can give away 4 e-. Transition metals are special in that they are able to give away many electrons.
So the number of O atoms dictates the number of metal atoms, M, that we need, but the number of M atoms needed also depends on the charge that the metal ion can make.
So for any metal oxide, we can assign the general formula MxOy the total negative charge needed to be balanced by the metal is 2y (each O has a double charge).
So for x metal atoms, the average charge on each metal ion must be:
(total negative charge ÷ No. of metal atoms)
(2y ÷ x)
Reaction of acid (abbreviated to H+) with metal oxides sees metal oxides to form hydrated metal ions and H2O.
You may be interested to know reaction of metal oxide with just water, produces hydroxides instead.
The O in from H2O comes from the O2- . For every O present in the metal oxide, 2 H+ are required. So the number of H necessary is also = 2y.
Al2O3 therefore needs 6 H+(aq) and produces 2 Al3+(aq) and 3 H2O(l)