Question

What molar quantity of `PbI_2` can be formed from a `5.85*mol` quantity of `"potassium iodide"`?

Answer 1

Approx. `2.93*mol` `PbI_2`.

Explanation:

We need to interrogate the stoichiometric reaction:

`Pb^(2+) + 2I^(-) rarr PbI_2(s)darr`

If there are `5.85*mol` `KI` then at most we can AT most get `2.925*mol` `PbI_2`.

SO 2 questions:

what mass of `PbI_2` does this molar quantity represent;

what is the colour of `PbI_2`?