Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For neutralization of acetic acid, the unbalanced equations are
(1) CH₃CO₂H(l) + Na₂CO₃(s) → CH₃CO₂Na(aq) + CO₂(g) + H₂O(l)
(2) CH₃CO₂H(l) + NaHCO₃(s) → CH₃CO₂Na(aq) + CO₂(g) + H₂O(l)
(b) How many kilograms of each substance is needed to neutralize a 1.000-gallon spill of pure acetic acid (density = 1.049 g/mL)?

Question

Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For neutralization of acetic acid, the unbalanced equations are

(1) CH₃CO₂H(l) + Na₂CO₃(s) → CH₃CO₂Na(aq) + CO₂(g) + H₂O(l)

(2) CH₃CO₂H(l) + NaHCO₃(s) → CH₃CO₂Na(aq) + CO₂(g) + H₂O(l)

(b) How many kilograms of each substance is needed to neutralize a 1.000-gallon spill of pure acetic acid (density = 1.049 g/mL)?

Accepted Answer

Hey everyone, we're told that barium hydroxide or potassium hydroxide reacts with acids in neutralization reactions, the unbalanced equations are as follows, how many kg of each item are required to neutralize 1.50 gallons of hydrochloric acid with a density of 1.18 g per mil leader. So first let's go ahead and balance out our equations. Looking at our first equation, we have barium hydroxide plus our hydrochloric acid, and we produce our barium chloride, plus water. To balance this out, all we need to do is add a coefficient of two prior to hydrochloric acid and a coefficient of two prior to water. And if we count the number of atoms per element, we find that it is completely bounced out. Now, looking at our second equation, we can see that this is already balanced out as is so we do not need to add any coefficients onto this equation. Now that we've bounced out our equations, let's go ahead and determine the mole of hydrochloric acid. Starting off with 1.50 gallons of hydrochloric acid. We can go ahead and use our conversion factor and this is going to be per one gallon. We have 3.7854 L converting our leaders into milliliters. We know that per one leader, we have 10 to the third milliliters. Using our density of 1.18 g per milliliter. And finally using our molar mass of hydrochloric acid and we know that per one mole of hydrochloric acid, we have 36.458 g of hydrochloric acid. And when we calculate this out and cancel out our units, we end up with a value of 183. mole of hydrochloric acid. Now that we've calculated our mole of hydrochloric acid. Let's go ahead and calculate the kilograms of barium hydroxide required to neutralize our hydrochloric acid. Starting off with our 183. mole of hydrochloric acid. We're going to look at our multiple ratios between hydrochloric acid and barium hydroxide. Looking at our equation above, we can see that per two mol of hydrochloric acid, We have one mole of barium hydroxide. Next using barium hydroxide molar mass, We have 171.34 g of barium hydroxide per one mole of barium hydroxide. And finally, since we were asked to do this in kilograms, we can convert our grams into kilograms and we know that we have 10 to the third grams per one kg. So when we calculate this out, We end up with a value of 15. kilograms of barium hydroxide. Now let's go ahead and do this for potassium hydroxide Again, taking the same steps, we're going to start off with 183. mol of hydrochloric acid. Looking at our multiple ratios between hydrochloric acid. We can see that we have one mole of hydrochloric acid per one mole a potassium hydroxide, Looking at potassium hydroxide smaller mass, we can see that we have 56.106g of potassium hydroxide per one mole of potassium hydroxide. Again, we do want this in kg. So we know that we have 10 to the 3rd g per one kg. So when we calculate this out, We end up with a value of 10.3110 kg of potassium hydroxide. And these are going to be our final answers. Now, I hope that made sense and let us know if you have any questions.

Verified Solution

Key Concepts

Neutralization Reaction

Stoichiometry

Density and Volume Conversion