\(Fe\left(0,15\right)+2HCl\left(0,3\right)--->FeCl_2+H_2\)
\(m_{HCl}=\dfrac{21,9.50}{100}=10,95\left(g\right)\) \(\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\) Theo PTHH: \(n_{Fe}=0,15\left(mol\right)\) \(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\) Vậy....