We're asked to find the theoretical yield, in #"kg"#, of #"Fe"(s)# that can be produced by reacting #25.0# #"kg Fe"_2"O"_3(s)#.

We can solve this problem by first converting the given mass of #"Fe"_2"O"_3# to *moles* using its *molar mass*, calculated to be #159.69# #"g/mol"#. (We need to convert from kilograms to grams as well):

#25.0cancel("kg Fe"_2"O"_3)((10^3cancel("g"))/(1cancel"kg"))((1color(white)(l)"mol Fe"_2"O"_3)/(159.69cancel("g Fe"_2"O"_3))) = color(red)(156.56# #color(red)("mol Fe"_2"O"_3#

Using the coefficients of the chemical equation, we can find the relative number of moles of #"Fe"# produced:

#color(red)(156.56)cancel(color(red)("mol Fe"_2"O"_3))((2color(white)(l)"mol Fe")/(1cancel("mol Fe"_2"O"_3))) = color(green)(313.11# #color(green)("mol Fe"#

Lastly, we can use the molar mass of iron (#55.85# #"g/mol"#) to find the number of grams, and then kilograms, of #"Fe"# that can be produced:

#color(green)(313.11)cancel(color(green)("mol Fe"))((55.85cancel("g Fe"))/(1cancel("mol Fe")))((1color(white)(l)"kg")/(10^3cancel("g"))) = color(blue)(17.5# #color(blue)("kg Fe"#

rounded to #3# significant figures.