A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs: 2 KCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2 KNO₃(aq) The solid PbCl₂ is collected, dried, and found to have a mass of 2.45 g. Determine the the percent yield.

Video transcript

Welcome back everyone to another video, a 25 mL sample of a 1.20 molar potassium chloride solution is mixed with 15 mL of a 0.900 molar A lead two nitrate solution. And this precipitation reaction occurs two moles of potassium chloride react with one mole of lead two nitrate pro and they produce one mole of lead, two chloride and two moles of potassium nitrate. The solid lead two chloride is collected dried and found to have a mass of 1.13 g. Determine the percent yield. This is what we're looking for in this problem option. A states that the answer is 66.7% B 3.74% C, 30.1% and D 116%. Apparently, we can immediately say that D cannot be a correct option. We cannot have a percent yield greater than 100. So now let's solve this problem. First of all, we're given our equation and what we're going to do is just calculate the number of moles of each reactant because we have to determine the limiting reactant. So we're going to start with potassium chloride, the number of moles of potassium chloride can be found as the product of molarity and volume. So we're taking 1.20 molar and we have to multiply by the volume in liters, we have 25.0 mL. So that means we need to multiply by 10 to the negative third to get leaders. This is how we get the number of most of potassium chloride. Let's calculate it later. But now we just want to focus on the same approach and find the number of moles of lead to nitrate. We take its molarity 0.900 molar and we multiply by the volume in this case, 15 mL which is 15 multiplied by 10 to the negative third of a liter. Now, at this step, it's really important to calculate those values. And let's do that or the first valley, we end up with zero point 0300 moles. Now, for the second one, we get 0.0135 moles. Now we have to find the limiting reactant and it's not necessarily the lowest value. Now, what we have to do is just compare their equivalents. So potassium chloride has a coefficient of two. So that means we have to take the number of moles of KCL, we have to divide that by two, which essentially gives us 0.0150 moles, let two nitrate has a coefficient of one. So we can essentially divide 0.0135 by one, which gives us the same number. So now which one is smaller, well, essentially 0.0135 moles is smaller than 0.0150. Which means that let to nitrate is the limiting reactant. Now that we know our limiting reactant, we will find the theoretical number of moles of the product lead to chloride. According to ST geometry, we can say that one mole of the limiting reactant produces one mole of lead two chloride. So the theoretical number of moles of lead two chloride would be equal to 0.0135 moles. From here, we can find the theoretical mass of lead two chloride, taking the number of moles and multiplying by the molar mass of a lead to chloride. Now the molar mass of lead two chloride would be 278.1 g per mole. And the theoretical yield that we get is equal to 3.75 or grams. Let's add that extra significant figure. Now the percent yield is simply the ratio between the actual mass produced. So that would be 1.13 g. And we're going to divide that by 3.754 which is theoretically yield then multiply by 100 right. So mass actual divided by mass theoretical multiplied by 100. And we can essentially state that the percent yield would be 30.1% looking at the answer choices, this corresponds to the answer choice. C Thank you for watching.