Ammonium chloride, NH4Cl, is a very soluble salt in water. (d) How many grams of silver nitrate do you need to add to the solution in part (c) to precipitate all of the chloride as silver chloride?

Video transcript

Hi everyone today, we have a question telling us a 17.0 g sample of sodium chloride Was dissolved in 250 ml of water. And our goal is to calculate the grams of lead to nitrate needed to precipitate all of the chloride as lead to chloride. So the first thing we need is a balanced equation. So we have sodium chloride plus our lead to nitrate. And we're going to displace this. So our sodium with a plus one charge is going to go with our nitrate With a -1 charge. And our goal here is to have a charge of zero. So plus 1 -1 will cancel out. So we'll just have in a in 03 and then we're going to do the same with lead and chlorides. So Lead is going to have a plus two charge and chloride has a negative one charge. So we need to crisscross these to get a charge of zero. So we'll have P B C L two. So this is forming our sodium nitrate plus lead chloride. And now we need to balance this. So if we look at our reactive side, we see that we have to nitrates and on the right, we only have one. So we're gonna put a two in front of our sodium nitrate and now we have to sodium is on our product side and only one on our reactive side. So we're going to put a two in front of our sodium chloride and now it is all balanced. So now that we have a balanced equation, we can put this into an equation. So we're gonna start with 17 g of sodium chloride, times one more of sodium chloride over It's Mueller mass, which is 58 .44g of sodium chloride. And we're going to multiply that by our multiple ratio. So one more of lead nitrate over two moles of sodium chloride, times Our molar mass of lead nitrate, which is .22g over one mole of a lead nitrate. So our grams of sodium chloride are canceling out are moles of sodium florida, canceling out are moles of lead nitrate are canceling out, leaving us with grams of lead nitrate And that equals 0.2 g of lead nitrate. And that is our final answer. Thank you for watching. Bye.