When 75.0 mL of a 0.100 M lead(II) nitrate solution is mixed with 100.0 mL of a 0.190 M potassium iodide solu-tion, a yellow-orange precipitate of lead(II) iodide is formed. What is the mass in grams of lead(II) iodide formed? Assume the reaction goes to completion. (LO 4.11, 4.15)
(a) 1.729 g
(b) 3.458 g
(c) 4.380 g
(d) 8.760 g

Question

When 75.0 mL of a 0.100 M lead(II) nitrate solution is mixed with 100.0 mL of a 0.190 M potassium iodide solu-tion, a yellow-orange precipitate of lead(II) iodide is formed. What is the mass in grams of lead(II) iodide formed? Assume the reaction goes to completion. (LO 4.11, 4.15)
(a) 1.729 g
(b) 3.458 g
(c) 4.380 g 
(d) 8.760 g

Accepted Answer

Hello everyone. So in this video we want to calculate for the mass in grams of silver chloride precipitate that forms and we're assuming that the reaction goes to completion. Let's first go ahead and rewrite our right out our chemical reaction. So its first given to us in kind of just words here. But we'll go ahead and convert this into just purely chemistry. So our action then is when Our silver nitrate, which is a GN- 03, that is acquis that reacts with sodium chloride so that a cl. So we see here that we have a double displacement reaction. So that's when the an ionic and can ionic partners are being switched. So we have our A G C. L. As our first product which is a solid in case. And that's just our precipitate here. Our second product is N A N 03 and that is acquis. So this product here is what we're interested in and one thing to take you out of is that our molar mass of our precipitate here. So H. D C. L. That is equal to 1 43.22 g per mole. So let's go ahead and find our limiting reagents since we're given the information of two different storing materials. So starting off with our silver nitrate. So that's 50 mL of the solution. We want to convert our millions into leaders with a dual conversion. So for every 1000 ml, we have one leader now that we are in units of our leaders. We go ahead and utilize our concentration. So every 0.150 moles. Our silver nitrate goes into our one leader of our solution. So then leaders will cancel. Next. We can go ahead and do a multiple conversion of our certain materials with our product. So we have that for every one mole of our starting material here, We get one mole of our product, that's a gcl. So then we can see that the molds of our cinema tian is now canceled. And we're in terms of our precipitate. So last thing is to go ahead and utilize our molar mass that we said. So we get gramps. So for every one mole this maybe use the same color here. So for one mole of r a G C l, we have 143.22 g of a gcl. So you see now that the moles of hcl will cancel. And once we put everything into the calculator we get the value of 1.0749 g of our precipitate. Alright, now going to use our 2nd story material which is sodium chloride. So we're given 100 millions of this sodium chloride solution Again, let's go ahead and convert our Millimeter unit into leaders. So for every 1000 ml, we have one leader, the units of military councils, we can go ahead and use the concentration here. And that's saying that for every one leader of solution it contains 0.210 moles of sodium employed. So now our leaders units will cancel Now continue our dimensional analysis again, we can use a multiple ratio to get in terms of our precipitate. Everything here is a 1-1 ratio. So that makes it easier for us. So every one mole of sodium chloride, we get one mole our silver Floyd. Now we see that the moles of N A C L cancels. And lastly to get us into terms of grams and units of mass. We can go ahead and just use the molar mass. So for every one mole of a gcl we have 143.22 g of a gcl. So now we can see that the moles will cancel. Once I put everything into the calculator, I get the value of 3.0097 units being grams of sodium chloride. So now we can go ahead and compare these two masses. So we use silver nitrate, we get 1.0749 g of our precipitate. If we use unemployed, we get essentially 3g of our precipitate. So we see that this is definitely our smaller amount, which means that our silver nitrate is our limiting reagent. Since this is our Lebanon region, this is the actual amount of our precipitate that Formed. So that we can conclude from our calculations that's 1.07 g of our precipitate. That is silver chloride is formed. So then this right here is going to be my final answer for this problem.

Verified Solution

Key Concepts

Stoichiometry

Molarity

Precipitation Reactions