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Solution: Write balanced net ionic equations for the reactions that oc...

Question

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) Cr2(SO4)3 (aq) + (NH4)2CO3 (aq) →

Video Transcript

Here we're asked to figure out what the net ionic equation will be and also the spectator ions. So the compounds that we have reacting here are chromium 3 sulphate plus ammonium carbonate. So this is going to break up into its ions so it breaks up into chromium 3+ ion and sulphate ion. This breaks up into ammonium ion and carbonate ion. Now opposites attract, the 3 from here comes here, the 2 from here comes here so we get CR2CO33 and then here we base it off of the rules that we know for solubility and here carbonate here combined with this would form a precipitate so it'll be a solid plus 2 from here comes here, 1 from here comes here, anything with ammonium ion is going to be soluble so this is in parenthesis and put the little 2 there and then SO4. I've got to balance this out, we have 2, let's see to make sure everything's balanced, we have 2 ammoniums but then we have 3 sulphates here but only 1 here so we have to put a 3 here and a 3 here.

Now everything is balanced out. So for the complete or total ionic, we break up only aqueous compounds. Aqueous, aqueous and this is aqueous, these are the only ones that are going to break up. So we're going to have 2 chromium 3 ions, plus 3 sulphate ions plus 6 ammonium ions plus 3 carbonate ions produce, this is still a solid so it's still all together, plus 6 ammonium ions plus 3 sulphate ions. Your spectator ions are the ions that look the same on both sides of the equation which would be ammonium ions and sulphate ions. So those are your spectator ions. The net ionic is everything else left behind once you've taken out the spectator ions. So that would be the chromium 3 ions with the carbonate ions and now they combine together to give you your solid which is chromium 3 carbonate. So those are the steps you'd have to take in order to figure out your spectator ions and finally your net ionic equation.