Convert 10.0 g Pb to mols.
Convert 1.56 g S to mols.
Convert mols Pb and mols S to mols PbS, then convert to grams PbS for Pb AND for S. You will find that, within reasonable error, each will produce approximately the same amount of PbS.
Tnen do the same for 10 g Pb and 3.00 g S. You will find that the same amount of PbS is formed even though more S was used.
Then do the same for 18 g Pb. You will find the amount PbS is the same even though more Pb was used.
You can put all of that into your own words to explain the limiting reagent. Its Pb when more S is used but S when more Pb is used.
A researcher found that 11.56 g of lead sulfide, PbS, were formed when each of the following occurred:
10.0 g of lead reacted with 1.56 g of sulfur
10.0 g of lead reacted with 3.00 g of sulfur
18.0 g of lead reacted with 1.56 g of sulfur
Explain these observations in terms of the concept of limiting reactant.
1 answer
1 answer