A railroad tank car derails and spills 36 tons of concen-trated s...

Question

A railroad tank car derails and spills 36 tons of concen-trated sulfuric acid. The acid is 98.0 mass% H2SO4 and has a density of 1.836 g/mL. 
(b) How many kilograms of sodium carbonate are needed to completely neutralize the acid?

Accepted Answer

Welcome back everyone in a laboratory. A student accidentally spills 300 g of glacial acetic acid which is 99.5% acetic acid by mass, calculate the mass and grams of sodium bicarbonate required to neutralize the spill. Let's recall that sodium bicarbonate is represented by the formula N a H C 03. Where we would recognize that it's considered a weak base. Whereas our acetic acid we want to recognize is a weak acid. So reacting the two, we would have acetic acid, CH three C 02 H reacting with sodium bicarbonate, N A H C 03 to produce on our product side are conjugate base of our weak acid which is going to be sodium acetate. So that would be CH three C 02 N. A. And this is the salt of our conjugate base. We would also form carbon dioxide and water as our products. Since our acid is going to be donating a proton to form a bond with the hydrogen and oxygen atom in our sodium bicarbonate. So now that we have this reaction, let's make sure everything is balanced. And looking at all of our atoms on both sides of our equation are react inside and our product side, we can confirm that all of our atoms are actually balanced when you count them out. So we're just going to put a check mark for the fact that this equation is balanced as is with no needed additional coefficients. So what we're going to do now is take note of our mass of our acetic acid. So according to the prompt. So for our mass of ch three c 02 H we have a mass of 300 g and were given the percent by mass of acetic acid as 99.5% which we're going to interpret as decimal being 0.995. And so multiplying these two values, this gives us our massive acetic acid being 298. g. Now we want to make note of our molar mass of acetic acid from our periodic table, Where we will see that for every atom in acetic acid, we have a molar mass of 60.05 g per mole. We also want to make note of our molar mass of sodium bicarbonate from the periodic table of n a h c 03. And we would see that sodium bicarbonate corresponds to the molar mass of 84.1 g per mole. So now we're going to get into stock geometry to figure out our final answer which is our mass of sodium bicarbonate and grams required to neutralize the acetic acid spill. So beginning with what we know, we're going to begin with that mass, we calculated of 298.5 g of our acetic acid, ch three c 02 H. We're going to go from grams of acetic acid, two moles of acetic acid. By utilizing our molar mass, which we noted down is 60.5 g for one mole of acetic acid, allowing us to cancel out grams of acetic acid so far. And sorry, this is grams over here. So now we're at moles of acetic acid. We want our final answer to end up with sodium bicarbonate. So just to make more room here and let's fix this just to make more room. Our next conversion factor is to use our bounced equation to go from moles of acetic acid, two moles of sodium bicarbonate. Based on our balanced equation, we see we have a molar ratio where we have a coefficient of one in front of sodium bicarbonate and a coefficient of one in front of our acetic acid. So we have a 1-1 molar ratio which allows us to go from moles of acetic acid to now moles of sodium bicarbonate. Where we need to end up with our mass and grams of sodium bicarbonate. So continuing the calculation below, we're going to utilize that molar mass of our sodium bicarbonate. Where we see that we have from going from moles of sodium bicarbonate in the denominator. So it can cancel out two g of sodium bicarbonate. We have an equivalent of 84.01 g for one mole of sodium bicarbonate. Because our moles of sodium bicarbonate are aligned diagonally. We can cancel them out. And for our final unit, we're left with grams of sodium bicarbonate, which is going to give us our final answer of our Weak base, needed to neutralize our weak acid acetic acid of 417.6 g of sodium bicarbonate needed. So this is needed to neutralize the spill, and this would be our final answer to complete this example. So I hope this was helpful if you have any questions, leave them down below and I'll see everyone in the next practice video.

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