nFe=56/56=1(mol)
Fe+HCl=>FeCl2+H2
a, Theo PTPƯ: nFeCl2=nFe=1(mol)
=>mFeCl2=1.127=127(g)
b,theo ptpu ta có nH2=nFe=0,1(mol)
=>vH2(đktc)=1.22,4=22,4l
VH2(đkc)=1.24,79=24,79l
ko bt tính 25 độ 1 bar
\(a)n_{Fe}=\dfrac{56}{56}=1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{FeCl_2}=n_{H_2}=n_{Fe}=1mol\\ m_{FeCl_2}=1.127=127g\\ b)V_{H_2}=1.24,79=24,79l\)