During a certain time period, 4.0 million tons of SO₂ were released into the atmosphere and subsequently oxidized to SO₃. As explained in the Inquiry, the acid rain produced when the SO₃ dissolves in water can damage marble statues: CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + CO₂(g) + H₂O(l) (a) How many 500 pound marble statues could be damaged by the acid rain? (Assume that the statues are pure CaCO₃ and that a statue is damaged when 3.0% of its mass is dissolved.)

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Welcome back everyone of that containing five kg of acetic acid was accidentally spilled over a number of marble blocks. Acetic acid reacts with calcite in marble. By the following reaction. Where one mole of calcite reacts with one mole of acetic acid to produce one mole of calcium, acetate, one mole of car dioxide gas and one mole of liquid water. If the marble blocks have a mass of 100 g, determine the number of blocks that were damaged, assume the blocks are made up of pure calcite and it's considered damaged if 5% of the mass reacts. So our first step is to make sure we have a bounced equation. We have our calcium carbonate reacting with acetic acid to produce our product calcium acetate, one mole of carbon dioxide gas and one mole of liquid water. So making sure things are balanced, we're going to count our atoms on both sides of our equation, we have calcium, carbon, oxygen and hydrogen to worry about on both sides. So for our calcium atoms we can count one on both sides of our equation for carbon on the reactant side, we can count 123 On our reactant side. For carbon on the product side, we can count to four And then we have five on the product side. For oxygen on the reactant side, we can count three 45. For oxygen on the product side, we can count 4567. And for hydrogen on the reactant side we can count 34. For hydrogen on the product side, we can count six And then 7, 8. So now making sure things are bounced. We're gonna begin by placing a coefficient of two in front of our acetic acid on the reactant side, which is going to change our carbon atoms to now we would have to and then we would have four and then we would have five on the reactant side. We would also change our hydrogen to now six and then 7, 8 on the reactant side. And this is also going to again change our oxygen to now four and then 5, 6, 7 on the reactant side. And so now we can see that with that coefficient in front of our reactant acetic acid, we now have a balanced equation. Right now, we need to take note of our molar mass of our calcium carbonate. We would see that on our periodic table, calcium carbonate has a molar mass of 100.09 g per mole. And now for our molar mass of our acetic acid, We would see on our periodic table that it has a molar mass of 60.05 g Permal. And now calculating for our mass of calcium carbonate that is reacted which is really just our calcite marble reacting with the spilled acetic acid. We're going to begin with the amount of our acetic acid that is spilt, Which according to the prompt is five kg of acetic acid where we're going to multiply by the molar mass of acetic acid. Or rather we want to get rid of those units of kilograms. So we're actually going to multiply by our conversion factor to go from kilograms to grams by recalling that are prefix kilo, tells us that we have 10 to the third power of our base unit grams for one kg. We can now get rid of kilograms and now going from grams of acetic acid. We want to multiply by our next conversion factor to go from grams of acetic acid. two moles of acetic acid, which is finally where we use our molar mass. That we noted down from the periodic table where we saw that we have 60.5 g of acetic acid for one mole of acetic acid allowing us to get rid of grams of acetic acid and now we want to get from moles of acetic acid, two moles of our calcium carbonate. So we're going to continue our study geometry below where we're going to multiply by our molar ratio according to our bounced equation between moles of acetic acid and the denominator and then molds of our calcium carbonate in the numerator So that everything can cancel out accordingly. So looking at our bounced equation, we would note the fact that we have a molar ratio between our calcium carbonate and acetic acid, which is a 1-2 molar ratio. And so we would say we have one mole of calcium carbonate for two moles of acetic acid. So now canceling out our moles of acetic acid. We want to end up with the mass of calcium carbonate that reacts with the acid. So we're going to plug in our molar mass of acetic of, sorry, calcium carbonate from our periodic table where we said we have 100.9 g of calcium carbonate equivalent to one mole of calcium carbonate, allowing us to cancel out moles of calcium carbonate and leaving us with the mass of calcium carbonate that reacts with the acid equal to a value of 4166.94 g. And now we need to figure out based on this mass of calcium carbonate, how many of our calcite blocks which are made up of calcium carbonate are damaged. And so first we need to go back to the prompt and take note of the fact that According to the promise we have 5% of our massive calcium carbonate that is reacted. And so we have the marble blocks with a mass of 100 g. So we're going to see what mass of damage we have per block Where we take that 5% as a decimal being 0.05. And we're going to multiply by the 100 g per block of our calcium carbonate or calcite. So per block of calcite will say. And so let's make these parentheses neater. So multiplying these two values, we're going to yield a mass of damage per block, equal to five gramps of damage per block of calcite. And now for our final answer, we need to figure out how many blocks of calcite are damaged. And so we're going to take that mass of our calcium carbonate, which makes up our calcite blocks that reacted, which we determined is this value here. So 4166.94 g of our calcium carbonate that is reacted. We're going to multiply this by our damage per block. So we're going to use that as a conversion factor where in our denominator we plug in, we have five g of damage of calcium carbonate per one block of health site. And now we're able to cancel our units of grams of calcium carbonate, leaving us with blocks of calcite as our final unit, where we will determine based on our store geometry That we have a result of 833 blocks that are damaged from the acetic acid spill. And this would be our final answer. To complete this example. I hope everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video

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Key Concepts

Stoichiometry

Acid-Base Reactions

Mass Percent and Dissolution