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

1 Answer
May 1, 2017

Answer:

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#?