\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH : Fe + 2HCl -> FeCl2 + H2
0,2 0,4 0,2
Xét tỉ lệ \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => Fe đủ , HCl dư
\(m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\)
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
pthh:
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(nFe=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(nHCl=0,5\left(mol\right)\)
Xét tỉ lệ : \(\dfrac{nFe}{1}< \dfrac{nHCl}{2}\left(\dfrac{0,2}{1}< \dfrac{0,5}{2}\right)\)
suy ra : HCl dư
\(nHCl_{\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)
\(mHCl_{\left(dư\right)}=0,05.36,5=1,825\left(gam\right)\)
Vì HCl dư nên ta lấy số mol của Fe làm chuẩn
\(\Rightarrow nH_2=nFe=0,2\left(mol\right)\)
\(nH_{2\left(đktc\right)}=0,2.22,4=4,48\left(lít\right)\)
