A 55.0-mL sample of a 0.102 M potassium sulfate solution is
mixed...

Question

A 55.0-mL sample of a 0.102 M potassium sulfate solution is
mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and
this precipitation reaction occurs:
K2SO4(aq) + Pb(C2H3O2)2(aq)¡2 KC2H3O2(aq) + PbSO4(s)
The solid PbSO4 is collected, dried, and found to have a mass of
1.01 g. Determine the limiting reactant, theoretical yield, percent yield.

Accepted Answer

welcome back everyone in this example. We have 60 mL of 600.213 molar sodium sulfate solution and mL of 1.71 moller calcium acetate solution, which reacts according to the given reaction below. So we have one mole of sodium sulfate reacting with one mole of calcium acetate to produce two moles of sodium acetate and one mole of calcium sulfate solid. And we need to identify our limiting reactant, theoretical yield and percent yield, sorry, percent yield if 1.53g of our calcium sulfate precipitate is obtained. So we should recognize first that we are given a balanced reaction. So that's important to verify because our first step is going to be calculating our mass of our calcium sulfate that is produced by each of our reactant here and so beginning with our first reactant we want to make note of again our molar mass of our calcium sulfate, which we can find from our periodic tables and we see that that is equal to a value of 136.15 g per mole for every atom in calcium sulfate. So we want to figure out how much calcium sulfate we get first from our first reactant, sodium sulfate. And so to get that we're going to begin with our concentration of sodium sulfate given from the prompt as 0.213 Molar. But we want to recall that polarity can be interpreted in units of moles per liter. And so we're going to write this out as .2 and three moles divided by leaders. And this allows us to multiply by our volume given in the prompt which is actually given in ml here. And so we want to recall that we're going to have to convert from male leaders to leaders. And so we want leaders as our final unit and male leaders in the denominator. So it cancels out and recall that our prefix milli tells us that we have 10 to the negative third power of our base unit leader for one male leader. Now we're able to cancel our units of middle leaders leaving us with leaders as our final unit. And this gives us a volume of 0. L. So we're going to multiply by that value 0.06 L and last we want to go ahead and plug in that molar mass. So we're going to multiply by 1 36.15 g per mole. We want mole in the denominator because we want mass as our final unit in grams. So right now we can go ahead and cancel our units of leaders as well as moles. Leaving us with grams as our final unit. And in our calculators. This is going to yield an amount of 1.74 g of our calcium sulfate produced from sodium sulfate which is our first reacted. We want to compare this amount to the mass of our calcium sulfate produced from our second reactant being our calcium acetate. So C. Two H 3022. And so we're going to follow the same process. We're given the polarity for calcium acetate as 1.71 moller which we will interpret as 1.71 moles per liter. Because we recall that polarity is equivalent to moles per liter. We're going to multiply by our volume. So because it's 50 mL following the same process where we can converted 60 mL to liters, we're going to also multiply this by 10 to the negative third power. And that's going to give us 0.5 liters which we will then multiply by again our molar mass of our product calcium sulfate. So that's 1 36.15 g per mole canceling out our units. We can get rid of leaders, we can get rid of moles and again we're left with grams as our final unit. And this is going to yield an amount of 11. g of our calcium sulfate produced from our calcium acetate. Now that we have these two amounts determined, we can see that 1.74 g is less than 11.64 g. And so therefore we can say the reactant that produces the smaller amount which is our 1.74 g is going to be our limiting reactant. And so that means that therefore our sodium sulfate is our limiting reactant. So we'll use LR for short limiting reactant and this would be our first answer here and now our next part is finding the theoretical yield. And we should recall that once we know our limiting reactant, we know our theoretical yield because the limiting reactant determines your theoretical yield. And so because sodium sulfate is the limiting reactant, we would say our theoretical yield which will use T Y. For short, Is going to be 1.74 g of our calcium sulfate. So this is our second answer here. We can say thus. Our theoretical yield is 1.74 g. And now we want to find our percent yield based on 1.53 g of our solid calcium sulfate being obtained. And so to calculate our percent yield, We're going to take that value. Specifically the smaller value 1.53 g of our calcium sulfate That is obtained. And we're going to divide by our larger value which is our limiting or our theoretical yield of our calcium sulfate which we determined to be 1. g of calcium sulfate. And so this is going to give us our percent yield. And we have to also multiply by 100. So let's fit that in there. Multiplied by 100. We're going to have a percent yield equal to a value of 87.9 of our calcium sulfate that should be obtained. And so this is going to be our 3rd final answer. So everything highlighted in yellow represents our final answers. To complete this example, I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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