a)
Nguyên tử khối trung bình của iron:
\(\overline{M}=\dfrac{54.5,8+56.91,72+57.2,2+58.0,28}{100}=55,9116\)
b)
\(n_{Fe}=n_{FeS_2}=\dfrac{589,97932}{120,12}=4,9\left(kmol\right)\\ \Rightarrow n^{56}_{26}Fe=91,72\%.4,9=4,49428\left(kmol\right)\\ \Rightarrow m^{56}_{26}Fe=4,49428.56=251,68\left(kg\right)\)