Gas Evolution Equations (Simplified) - Online Tutor, Practice Problems & Exam Prep

Gas Evolution Equations (Simplified) Concept 1

A gas evolution equation is a molecular equation that involves the creation of carbon dioxide gas. Here we're going to say that gas is formed once medium products lose a water molecule. Now what do we mean by medium product? A medium product is just a form of product held before it fully converts into its final product form. So here, our final product for this type of reaction equals your median product minus water. So if we take a look here, the reactant ions you have to be on the lookout for when it comes to our gas solution equations is H⁺ reacting with our bicarbonate ion. When it does, it's going to create as our median product, carbonic acid. Now here's the thing, carbonic acid cannot be held as a product. It'll automatically become a final product after losing water. So subtract out water from this structure, and you'll see what you have left. You would have CO₂ after you remove the water. Now here, H⁺ carbonate ion would give us the same result because here it also makes carbonic acid as a median product. Again, subtract out water, and what you have left is CO₂. Now in actuality, you subtract out the water, but the water is still there. What's happening is this carbonic acid is splitting apart, and it's going to make a gas. So that's where the gas evolution equation comes from. You make CO₂ as a gas and water as a liquid.

Gas Evolution Equations (Simplified) Example 1

Here in this example question, it says, "predict whether a reaction occurs and write the balanced molecular equation." So here we have sodium carbonate reacting with hydrobromic acid. Alright. So step 1, we're going to break up reactant 1 and reactant 2 into their ionic forms. Sodium carbonate is composed of sodium ion, it's in group 1A, so it's +1, and carbonate ion, which is a polyatomic ion, so it's CO₃^2-. Hydrobromic acid is composed of H⁺ ion, H⁺1. Bromine is in group 7A, so it's -1. Step 2, 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 2 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 criss-cross. The 2 from here comes here, and the 1 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 3 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 acid which we know will break down further into water, which will be a liquid, plus CO₂ gas. So you do not keep the carbonic acid there. It completely breaks down to give you those 2 pieces.

And step 4, 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 1. So I put a 2 in front of this. CO₃, so we have 1 carbon and 3 oxygens. We have 1 carbon, 2 oxygens, 3 oxygens. So there goes the CO₃ there. Hydrogens, we have 1 hydrogen here and 1 Br here. But on the product side, we have 2 Br's because the 2 gets distributed to the Br, and 2 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.

Predict the products formed from the following gas evolution equation.
 

• A. KCl, H₂CO₃
• B. CO₂, H₂O​​​​
• C. KCl, CO₂, H₂O
• D. KCl, H₂, CO₃^2–