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

1 Answer
May 1, 2017

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?