Here in this example question, it says predict whether a reaction occurs and write the balanced molecular equation O. Here we have sodium carbonate reacting with hydrobromic acid. All right, so step one, we're going to break up reaction one and reactant 2 into their ionic forms. Sodium carbonate is composed of sodium ion. It's in Group 1A, so it's plus one, and carbonate ion which is a polyatomic ion, so it's CO32- hydrobromic acid is composed of H ion H1 Bromine is in Group 7A, so it's -1.
Step two, we swap ionic partners by remembering that opposite charges attract. So we're going to apply the rules that we've learned when it comes to creating new compounds from their ionic forms. If we take a look, this positive and this negative are attracted to one another, so they're going to combine. And remember, when your numbers and your charges are the same, the two ions just combine together and the charges disappear. So this will give us NaBr plus then we have here the numbers and the charges are different. When they're different, they don't just disappear and combine, they crisscross. The two from here comes here and the one from here comes here. That would give us, at this point H2CO3.
Now remember our solubility rules. NaBr Na is a group 1A ion. Anything connected to a group 1A ion is aqueous and soluble. And then look, we made H2CO3 that is one of our median products. So step three says identify the median product or gas that forms from the gas evolution equation except for hydrogen sulfide. Break it up into water and gas. So here we have carbonic acids which we know will break down further into water which will be a liquid plus CO2 gas South. You do you do not keep the carbonic acid there, it completely breaks down to give me those two pieces.
Step four. Last step says if necessary balance your molecular equation by placing the correct coefficients in front of each molecule. So here we just have to see is it balanced? We have here 2 sodiums and here we only have one, so I'd put a two in front of this CO3. So we have one carbon and three oxygens. We have one carbon 2 oxygens, 3 oxygens. So there goes the CO3 there hydrogens. We have one hydrogen here and one BR here, but on the product side we have two BRS because the two gets distributed to the BR and two hydrogens. So we'd have to put a 2 here. Now our equation is balanced, so these would be all the coefficients for this balanced gas evolution equation.