\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_S=\dfrac{6.4}{32}=0.2\left(mol\right)\)
\(Fe+S\underrightarrow{t^0}FeS\)
\(0.1........0.2\)
\(\dfrac{0.1}{1}< \dfrac{0.2}{1}\) \(\Rightarrow Sdư\)
\(m_{Cr}=m_{FeS}+m_{S\left(dư\right)}=0.1\cdot88+\left(0.2-0.1\right)\cdot32=12\left(g\right)\)