Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If 5.00 g of sulfuric acid and 5.00 g of lead(II) acetate are mixed, calculate the number of grams of sulfuric acid present in the mixture after the reaction is complete.

Video transcript

Welcome back everyone in this example, we're told that silver nitrate in solution form reacts with sodium chloride solution to give a white precipitate of silver chloride. We're told if 6.5 g of silver nitrate is mixed with 4.5 g of sodium chloride, how much sodium chloride will be left If all the silver chloride is precipitated. So our first step is going to be writing out our reaction. And according to the prompt, we have silver nitrate. So a GN- 03 would recall is our formula for silver nitrate. This is going to be reacting with sodium chloride. So, based on our cell ability roles, these are going to be soluble in water. So that's why they have the Aquarius label. And then on our product side, we have our precipitate of silver chloride. So that's a gcl and it gets a solid label because it's a precipitate. And then we also form our second product, according to the prompt being silver nitrate. So, N A N 03 and that is also soluble based on our sustainability rules that we should recall for nitrates. And what this question is essentially asking us is how much sodium chloride we have in excess. So how much is left over. So our first step is to find out the amount of moles we have of each of our reactant. So beginning first with our moles of our silver nitrate, We would take the mass given from the prompt as 6.5 g. And we're going to use a dimensional analysis step where we recall the molar mass of silver nitrate from our periodic tables, which equals 169.87 g of silver nitrate. And this is equivalent to one mole of silver nitrate. So this allows us to cancel out our units being grams of silver nitrate. We're left with moles of silver nitrate and sorry about that. So we're left with moles of silver nitrate. And this is going to give us a value equal to 0.0383 moles of our silver nitrate that is reacting or that we initially have at least Now we have our moles of our sodium chloride to calculate. And according to the prompt, we have a mass of 4.5 g of our sample of sodium chloride. We're going to multiply in a dimensional analysis step. Our molar mass of sodium chloride, which we recall in our denominator is 58.44 g of sodium chloride. And this is for one mole of sodium chloride. So what we wanna do is cancel out our units of our grams of sodium chloride. We're left with moles of sodium chloride. And this is going to give us a result equal to 0.0770 moles of our sodium chloride that we begin with. And because we see that our first reactant, the silver nitrate is a lower amount of molds only being this amount here. This means that therefore this is going to be our limiting reactant. So this is the actual amount of silver nitrate that is going to react and be used up in our reaction. However, since we have more of this sodium chloride left over, this means that we will have sodium chloride left over. And this is what we need to figure out how much is left over. So what we're going to do to figure that out is figure out how much moles of sodium chloride are actually reacting. So to figure out the moles of sodium chloride that react, we're going to look at the ratio between our limiting reactant silver nitrate and our sodium chloride. And we would get that by making sure we have a balanced equation and looking at our coefficients of our reactant. So looking at our equation, we can definitely confirm that it is all balanced based on our number of moles of reactant and products on both sides of our equation. And so we can see that we have a coefficient of one in front of silver nitrate and a coefficient of one in front of sodium chloride. And that means we have a 1-1 molar ratio. So to figure out our moles of sodium chloride that are reacting, we're going to take the moles of our silver nitrate, which we calculated above as 0.0383 moles of silver nitrate. And we're going to multiply by our multiple ratio between silver nitrate and sodium chloride and we said that we have one mole of our silver nitrate for one mole of our sodium chloride. This comes from our bounced reaction and this allows us to go from moles of silver nitrate, two moles of sodium chloride. So we're going to cancel out our moles of silver nitrate. And this gives us our moles of sodium chloride equal to a value Of 0.0383 moles of sodium chloride that actually are reacting. And now that we know this information, we can easily figure out how much sodium chloride we have left over by taking the difference. So we would say that our moles of sodium chloride remaining Is going to be the mass or the moles of sodium chloride that we start off with which above we calculated as 0.0770 moles of sodium chloride. This was our initial value. So we'll write initial And then we're going to subtract this from our moles of sodium chloride that we know react. Which above we calculated as 0.0383 moles of sodium chloride that are actually reacting. And this difference in our calculators will give us a value equal to 0.0387 moles of our sodium chloride. However, we need to figure out our mass of sodium chloride remaining in g. So we're going to just make more room here now that we have more room. We're going to multiply to go from moles to grams. So we're going to recall our molar mass of sodium chloride from the periodic table where we see that for one mole of sodium chloride, we have a mass of 58.44 g of sodium chloride. So now we're able to cancel our units of moles of sodium chloride. We're left with grams of sodium chloride And this is going to yield a final result left of our mass of sodium chloride being 2.26 g of sodium chloride. That remains. And this is, after all of our precipitate is formed. And so for our final answer, this is what is highlighted in yellow here are mass of our reactant sodium chloride that remains after our reaction. So I hope that everything I went through is clear. If you have any questions, leave them down below and I will see everyone in the next practice video.