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.
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!
1 answer
4 answers