Fe+2HCl->FeCl2+H2
0,25------------------0,25 mol
n H2=\(\dfrac{5,6}{22,4}\)=0,25 mol
=>m Fe=0,25.56=14g
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=5,6:22,4=0,25\left(mol\right)\)
\(n_{Fe}=0,25.1=0,25\left(mol\right)\)
\(m_{Fe}=0,25.56=14\left(g\right)\)
