Question

If you react 3.1g of lead nitrate with 2.7g of potassium iodide, what is the theoretical yield of the lead iodide precipitate?

Answer 1

The reaction here is:

Pb(NO₃)₂(aq) + 2KI(aq) $\rightarrow$ PbI₂(s) + 2KNO₃(aq)

Thus we can see that 1 equivalent of lead nitrate reacts with 2 equivalents of potassium iodide to yield 1 equivalent of lead iodide precipitate.

Given, 3.1 g of lead nitrate = 3.1g/331.2g/mol = 9.36*10^-3 moles of lead nitrate (mol. weight of lead nitrate is 331.2 g/mol)

And, 2.7 g of potassium iodide = 2.7g/166g/mol = 16.26*10^-3 moles of potassium iodide (mol. weight of potassium iodide is 166 g/mol)

Thus the ratio of lead nitrate to potassium iodide = 9.36:16.26 = 1:1.737

That means the amount of potassium iodide taken is less as compared to the chemical equation and it is the limiting reagent here. Thus the moles of Lead iodide expected to form = 16.26*10^-3/2 moles = 8.13*10^-3 moles of lead iodide.

Now the molecular weight of lead iodide is = 461.01 g/mol. Thus theoretical yield of precipitate lead iodide = 461.01 g/mol*8.13*10^-3 moles = 3.748 g.