A 56.0mL sample of a 0.102 M potassium sulfate solution is mixed with 34.7mL of a 0.114 M lead acetate solution and this precipitation reaction occurs:
K2SO4(aq)+Pb(C2H3O2)2(aq)→PbSO4(s)+2KC2H3O2(aq) The solid PbSO4 is collected, dried, and found to have a mass of 1.16 g.
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Determine the theoretical yield.
Determine the percent yield.
find the moles you started with K2SO4, nad moles started with Pbacetate.
the mole ratio in the equation is 1:1, so which ever mole started with is lowest, that is the limiting reactant., and you will get the same number of moles of lead sufate. Change that to grams, that is the theoritical yield
percent yeld= 1.16/theoyield * 100
Is this correct?
K2SO4 0.00573
Pbacetate 0.00307
limiting reactant is Pb(C2H3O2)2.
Grams lead sufate. 303.26g
Is theoretical yield 303.26g?
Percent Yield= 1.16g/303.26g * 100= 38 %
Some parts right; some wrong.
I didn't get the same answer you did when multiplying M x L = ?
For mols K2SO4 I obtained 0.00571 and for
Pb(ac)2 I obtained 0.00396. That doesn't change the identification of the limiting reagent but it points up a problem on using your calculator.
Then for grams PbSO4 that 303.25 is the molar mass. You find grams by mols x molar mass. mols = 0.00571
0.00571 x 303.25 = about 1.7 (you need it more accurately) and that is the theoretical yield.
Then % yield = (1.16/1.7)*100 = ?
While I'm at it let me point out another calculator problem. (1.16/303.26)*100 = 0.38% and not 38%.
Thats wrong!
The correct answer:
Determine the theoretical yield.
1.195 g
Determine the percent yield.
97.0 %
Well, to determine the theoretical yield, we need to calculate the moles of the limiting reagent, which is the one that will be completely used up in the reaction.
First, let's calculate the moles of K2SO4:
moles of K2SO4 = concentration × volume
= 0.102 M × (56.0 mL/1000 mL) (converting mL to L)
= 0.005712 moles
Next, let's calculate the moles of Pb(C2H3O2)2:
moles of Pb(C2H3O2)2 = concentration × volume
= 0.114 M × (34.7 mL/1000 mL) (converting mL to L)
= 0.003964 moles
Now let's determine the limiting reagent. Since the reaction requires 1 mole of K2SO4 for every 1 mole of Pb(C2H3O2)2, we can see that the reaction will stop when all the K2SO4 is used up. So, K2SO4 is the limiting reagent.
Next, let's calculate the moles of PbSO4 formed. Since the molar ratio between PbSO4 and K2SO4 is 1:1, we know that the moles of PbSO4 formed will also be 0.005712 moles.
Lastly, let's calculate the theoretical yield:
mass = moles × molar mass of PbSO4
= 0.005712 moles × 303.26 g/mol (molar mass of PbSO4)
= 1.7338 g
So the theoretical yield is 1.7338 grams.
Now, to determine the percent yield:
percent yield = (actual yield/theoretical yield) × 100
The actual yield given in the problem is 1.16 grams, so the percent yield is:
percent yield = (1.16 g/1.7338 g) × 100
= 66.9%
Therefore, the percent yield is approximately 66.9%.
To determine the theoretical yield and percent yield, we need to use stoichiometry to calculate the amount of PbSO4 that should have formed, and then compare it to the actual amount obtained.
Let's start by balancing the equation to determine the molar ratio between reactants and products:
K2SO4(aq) + Pb(C2H3O2)2(aq) → PbSO4(s) + 2KC2H3O2(aq)
From the balanced equation, we can see that 1 mole of PbSO4 is produced for every mole of Pb(C2H3O2)2 reacted.
Next, we need to determine the number of moles of PbSO4 that should have formed, based on the given volume and molarity of the lead acetate solution.
Number of moles of Pb(C2H3O2)2 = volume (in liters) x molarity
= 0.0347 L x 0.114 mol/L
= 0.003960 mol (rounded to 4 significant figures)
Since the stoichiometric ratio is 1:1 between PbSO4 and Pb(C2H3O2)2, we also have 0.003960 mol of PbSO4.
Next, we need to calculate the mass of PbSO4 that should have formed using its molar mass.
Molar mass of PbSO4 = atomic mass of Pb + atomic mass of S + (4 x atomic mass of O)
= (207.2 g/mol) + (32.1 g/mol) + (4 x 16.0 g/mol)
= 303.3 g/mol
Theoretical yield of PbSO4 = number of moles x molar mass
= 0.003960 mol x 303.3 g/mol
≈ 1.20 g (rounded to 2 significant figures)
Therefore, the theoretical yield of PbSO4 is approximately 1.20 g.
To calculate the percent yield, we will use the following formula:
Percent yield = (actual yield / theoretical yield) x 100
The actual yield is given as 1.16 g.
Percent yield = (1.16 g / 1.20 g) x 100
≈ 96.7% (rounded to 3 significant figures)
Therefore, the percent yield is approximately 96.7%.