Question

Answer 1

\(n_{FeO}=\dfrac{2,16}{72}=0,03mol\\ n_X=\dfrac{0,224}{22,4}=0,01mol\\ \overset{+2}{Fe}O\underrightarrow{\overset{+5}{+HNO_3}}\overset{+3}{Fe}\left(NO_3\right)_3+\overset{z}{X}+H_2O\\ BTe:n_{FeO}=\left(5-Z\right)n_X\\ \Leftrightarrow0,03=\left(5-Z\right)0,01\\ \Leftrightarrow Z=+2\\ \Rightarrow X=NO\)