The key here is the fact that #1# mole of any element contains #6.022 * 10^(23)# atoms of that element, as given by Avogaro's constant.
So if your sample contains #1# mole of iron, it contains #6.022 * 10^(23)# atoms of iron. Similarly, if a sample contains #1# mole of potassium, it contains #6.022 * 10^(23)# atoms of potassium.
Now, you know that iron has a molar mass of #"55.845 g mol"^(-1)#. This means that every time your sample contains #1# mole of iron, it will have a mass of #"55.845 g"#. This means that #6.022 * 10^(23)# atoms of irion, the equivalent of #1# mole of iron, havea mass of #"55.845 g"#.
Potassium has a molar mass of #"39.0983 g mol"^(-1)#, which means that #1# mole of potassium has a mass #"39.0983 g"#. This means that #6.022 * 10^(23)# atoms of potassium have a mass of #"39.0983 g"#.
So, use this to find the number of atoms of potassium present in #"780 g"# of potassium.
#780 color(red)(cancel(color(black)("g"))) * (6.022 * 10^(23) quad "atoms K")/(39.0983 color(red)(cancel(color(black)("g")))) = 120.137 * 10^(23) quad "atoms K"#
Now all you have to do is to figure out how many grams of iron will contain #120.137 * 10^(23)# atoms of iron.
#120.137 * color(blue)(cancel(color(black)(10^(23)))) color(red)(cancel(color(black)("atoms Fe"))) * "55.845 g"/(6.022 * color(blue)(cancel(color(black)(10^(23)))) color(red)(cancel(color(black)("atoms Fe")))) = color(darkgreen)(ul(color(black)("1100 g")))#
The answer is rounded to two sig figs, the number of sig figs you have for the mass of iron.
Notice that because #1# mole of atoms of iron is heavier than #1# mole of atoms of potassium, which is essentially saying that #1# atom of iron is heavier than #1# atom of potassium, the mass of iron that contains the same number of atoms as the number of atoms of potassium present in #"780 g"# of potassium is #># #"780 g"#.
