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Ch.4 - Chemical Quantities & Aqueous Reactions

Chapter 4, Problem 26

Consider the unbalanced equation for the neutralization of acetic acid: HC2H3O2(aq) + Ba(OH)2(aq) → H2O(l) + Ba(C2H3O2)2(aq) Balance the equation and determine how many moles of Ba(OH)2 are required to completely neutralize 0.461 mole of HC2H3O2.

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Hey everyone. Our question here is asking us to determine the mass in g of calcium hydroxide required to completely neutralize 3.41 g of acetic acid. Now looking at our equation, we can see that it's unbalanced. Since we have two acetate ions connected to our calcium and only one acetate ion in our react inside. In order to bounce this out, we're going to have to add a coefficient of two prior to our acetic acid. And since we added this coefficient of two, we're going to need to recheck our hydrogen as well In order to balance out our hydrogen is we're going to need to add a coefficient of two before our H20, which is our water. So rechecking if everything is balanced, we have our complete balanced chemical equation. So let's go ahead and move on with our problem. So starting off with 3.41 g of acetic acid. We're going to need to get the molar mass of acetic acid. And when we look that up, we end up with a molar mass of 60.052 g per mole of acetic acid. So, plugging that into our dimensional analysis, We have 52 g per one mole of acetic acid. Now we want to relate this to moles of calcium hydroxide. So looking at our balanced chemical equation, we see that per two moles of acetic acid, we have one mall of calcium hydroxide. Now in order to figure out the grams, we're going to need to calculate the molar mass of calcium hydroxide. And when we do that, we find that our molar mass is going to be 74.09 g per mole of calcium hydroxide. And plugging this into our dimensional analysis, We know that we have 74.09 g of calcium hydroxide per one mole. So calculating this out, We end up with a value of 2.10 g of calcium hydroxide, and this is going to be our final answer. So I hope this made sense and let us know if you have any questions.
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