\(n_{NO} = \dfrac{3,136}{22,4}= 0,14\ mol\ ;\ n_{Fe(NO_3)_3} = \dfrac{2,7m}{242} (mol)\\ \Rightarrow n_{Fe(NO_3)_2} = \dfrac{m}{56} - \dfrac{2,57m}{242}(mol)\\ BT\ e\ : 2(\dfrac{m}{56} - \dfrac{2,7m}{242}) + 3.\dfrac{2,7m}{242} = 0,14.3\\ \Rightarrow m = 8,96\ gam\)