Solutions of sulfuric acid and lead(II) acetate react to form sol...

Question

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 lead(II) acetate present in the mixture after the reaction is complete.

Accepted Answer

Hey everyone in this example, we have hydrochloric acid solution reacting with silver nitrate to produce our products, silver chloride, which is a solid precipitate and nitric acid. We need to figure out the mass of hydrochloric acid RV agent that remains in solution. So we need to figure out our excess reactant leftover in grimm's. So we want to go ahead and make note of the molds of each of our re agents in our given reaction. And also there should be a three here next to the oxygen for silver nitrate. We also want to make sure that this equation is balanced and based on the total amount of atoms for each of our elements on both sides of our equation. This definitely is a balanced reaction. So, beginning with finding the moles of our hcl, we're told that according to this question, we use 15 g of hcl. So we're gonna start out with 15 g of hcl and we're gonna use we're gonna go from grams to moles of hcl. So we're going to use the periodic table to refer to our molar mass for hydrochloric acid, which we see is equal to a mass of 36.46 g for one mole of hcl. And this is going to give us 0.411 moles of our hcl re agent. We also want to find moles of our second re agent, which is our silver nitrate. And we are given the mass used as 10 g of silver nitrate from the prompt. So we want to go ahead and convert from grams of silver nitrate, two moles of silver nitrate. So sorry, that should be a three there. And this here says moles. So using our periodic tables, we would see that the molar mass for silver nitrate Is equal to a value of 169.87 g for one mole of silver nitrate. And so canceling out our units of g were left with moles of silver nitrate. Which we're going to get a value equal to 0.0589 moles of silver nitrate. And because we see that 0.0589 moles of silver nitrate Is less than 0. moles of hydrochloric acid. We can say that therefore hcl is left in access. So we have Hcl or hydrochloric acid left over after our reaction is complete. And so we need to find our moles of hydrochloric acid that is consumed to figure out how much is left over and excess and moles. So we're gonna take our .411 moles of our hydrochloric acid and subtract that from our second reactant, which we know is our silver nitrate. And we said above that we would Have 0. moles of silver nitrate reacting. And this difference gives us our moles of hydrochloric acid consumed equal to 0.352 moles. Or rather we can say that this is our moles of hcl that is remaining. And so now that we have this value in moles, we want to actually get this in grams to completely answer this question because it's asking for the mass of Hcl remaining in the solution. So we would say that the mass of Hcl that remains Is going to equal our moles which we just said leftover is 0.352 moles of hcl. And we want to go from moles of hcl two g of hcl. So we're going to utilize our periodic tables and find the molar mass of HcL which we can recall is equal to 36.46 g of Hcl for one mole of Hcl. Now we're able to cancel our units of moles leaving us with our massive hcl which is going to give us our final results equal to 12.8 g of hcl that is left over in excess and this is going to be our final answer to complete this example. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

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 lead(II) acetate present in the mixture after the reaction is complete.

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