Write balanced molecular and net ionic equations for the reactions of (d) aluminum with formic acid, HCOOH.
So in this reaction we have aluminium solid being dissolved in formic acid. So if we look on the list, we see that aluminium is up here, definitely can displace the hydrogen from the acid to produce H2 gas. Now remember, the hydrogen that's being displaced is the hydrogen that's on the oxygen. Remember hydrogens on carbon are incredibly difficult to remove, so it's that H that I highlighted that's coming off.
So then here, we're going to have aluminium which is in group 3A so it's +3 with the formate ion which is +1. Number are different so they criss-cross, so we have aluminium formate aqueous plus H2 gas being produced. Here we have to balance things because they're definitely not balanced so here if I put, let's see, we have 3 formates here but we technically only have 1 here. If I put a 3 here, what will that do? That'll give me 3 there, then that'll be, I'll need 1.5 here which is not possible so here we'll put a 6 here, we'll have to put a 2 here, we'll have to put a 3 here and a 2 here and now it's balanced. So that's our balanced equation. Now remember for the total or complete ionic, we have to break up only aqueous compounds. So I have 2 aluminium solids plus 6 formate ions plus 6 H+ ions.
We have 2 aluminium ions plus 6 formate ions again plus 3H2 gas. Remember for the net ionic we remove spectator ions. The spectator ions are the 6 formate ions so they don't get brought down. At the end we'll have 2 aluminium solids plus 6 H+ aqueous which gives me 2 aluminium ions plus 3 hydrogen gas molecules. So remember we use the activity series as long as the metal is above hydrogen it can displace hydrogen from the compound to produce H2 gas.