Kim and Al need to determine the actual mass of lead(II) sulfate in grams that formed when 1250.0mL of 0.0500 M lead(II) nitrate was mixed with 0.2000 kL of 0.0250 M sodium sulfate. When the experiment was completed Kim determined that they had a 67.5% yield.
1)How many grams of lead(II) sulfate did Kim and Al produced?
2)Write a balanced equation and net ionic equation for this reaction!
Pb(NO3)2 + Na2SO4 ==> PbSO4 + 2NaNO3
mols Pb = 1.250 L x 0.0500 M = 0.0625
mols Na2SO4 = 200 L x 0.0250 M = 5 mols.
Pb(NO3)2 is the limiting reagent and 0.0625 mols PbSO4 will be formed.
g PbSO4 theoretical yield = 0.0625 mols x 303.26 g/mol = 18.95 g theoretical yield.
18.95 x 0.675= ?? grams actual yiel
The molecular equation is written above.
You should be able to get the ionic equation from it. Explain what you don't understand if you have trouble with it.
Check my thinking.
To determine the grams of lead(II) sulfate produced, we need to follow a series of steps:
1) Calculate the moles of lead(II) nitrate (Pb(NO3)2) and sodium sulfate (Na2SO4) used:
Volume of lead(II) nitrate solution = 1250.0 mL = 1.2500 L
Volume of sodium sulfate solution = 0.2000 kL = 200.0 L
Moles of Pb(NO3)2 = volume (L) x concentration (M)
= 1.2500 L x 0.0500 M
= 0.0625 moles
Moles of Na2SO4 = volume (L) x concentration (M)
= 200.0 L x 0.0250 M
= 5.00 moles
2) Use the balanced equation to determine the stoichiometry between the reactants and the product. The balanced equation for the reaction is:
Pb(NO3)2 (aq) + Na2SO4 (aq) -> PbSO4 (s) + 2NaNO3 (aq)
3) From the balanced equation, we see that the mole ratio between Pb(NO3)2 and PbSO4 is 1:1. Therefore, the number of moles of PbSO4 produced is also 0.0625 moles.
4) Next, we need to convert the moles of PbSO4 into grams. The molar mass of PbSO4 is:
Molar mass of PbSO4 = (1 x atomic mass of Pb) + (1 x atomic mass of S) + (4 x atomic mass of O)
= (1 x 207.2 g/mol) + (1 x 32.1 g/mol) + (4 x 16.0 g/mol)
= 207.2 g/mol + 32.1 g/mol + 64.0 g/mol
= 303.3 g/mol
Mass of PbSO4 = moles x molar mass
= 0.0625 moles x 303.3 g/mol
= 18.95625 grams
5) Finally, we need to take into account the 67.5% yield. Yield is the ratio of the actual amount obtained to the theoretical amount predicted by stoichiometry.
Actual mass of PbSO4 = yield x theoretical mass
= 0.675 x 18.95625 g
≈ 12.79 grams
1) Kim and Al produced approximately 12.79 grams of lead(II) sulfate.
2) The balanced equation for the reaction is:
Pb(NO3)2 (aq) + Na2SO4 (aq) -> PbSO4 (s) + 2NaNO3 (aq)
The net ionic equation, which shows only the species actually participating in the reaction, is:
Pb²⁺ (aq) + SO₄²⁻ (aq) -> PbSO₄ (s)