Identify the limiting reagent, and determine how many moles of the excess reagent remain?

The balanced chemical equation that describes this synthesis reaction looks like this

#color(red)(2)"K"_ ((s)) + "I"_ (2(g)) -> 2"KI"_ ((s))#

The two reactants react in a #color(red)(2):1# mole ratio, so right from the start you know that the reaction consumes twice as many moles of potassium metal than moles of iodine gas.

You were given grams, so use the molar masses of the two reactants to go to moles

#15.0 color(red)(cancel(color(black)("g"))) * "1 mole K"/(39.1color(red)(cancel(color(black)("g")))) = "0.3836 moles K"#

#20.0 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.8color(red)(cancel(color(black)("g")))) = "0.07880 moles I"_2#

Notice that you have about fewer moles of iodine gas than would be needed to allow for all the moles of potassium to take part in the reaction.

This is the case because #0.07880# moles of iodine gas will only allow for

#0.07880color(red)(cancel(color(black)("moles I"2))) * (color(red)(2)color(white)(a)"moles K")/(1color(red)(cancel(color(black)("mole I"2)))) = "0.1576 moles K"#

to take part in the reaction. Therefore, iodine gas will be the limiting reagent. This of course implies that potassium is in excess.

The number of moles of potassium that remain unreacted will be equal to

#n_"K remaining" = "0.3836 moles" - "0.1576 moles"#

#n_"K remaining" = color(green)(|bar(ul(color(white)(a/a)"0.226 moles K"color(white)(a/a)|)))#

The answer is rounded to three sig figs.