\(PTHH:Fe+2HCl->FeCl_2+H_2\uparrow\)
\(a,^nFe=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(^nHCl=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
Theo PT và bài ra ta có: \(^nFe_{\left(dư\right)}\) Tính theo \(^nHCl\)
Theo PT: \(^nFe=\dfrac{1}{2}^nHCl=\dfrac{1}{2}0,5=0,25\left(mol\right)\)
\(^nFe_{\left(dư\right)}=0,5-0,25=0,25\left(mol\right)\)
\(^mFe_{\left(dư\right)}=56.0,25=14\left(g\right)\)
b, Theo PT: \(^nH_2=^nFe=0,25\left(mol\right)\)
\(V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)