\(n_{Fe}=\dfrac{1.12}{56}=0.02\left(mol\right)\)
\(n_{HCl}=0.1\cdot0.3=0.03\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.015.....0.03.......0.015\)
\(m_{Fe\left(dư\right)}=\left(0.02-0.015\right)\cdot56=0.28\left(g\right)\)
\(m_{FeCl_2}=0.015\cdot127=1.905\left(g\right)\)