\(n_{Fe}=\dfrac{0.56}{56}=0.01\left(mol\right)\)
\(n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có :
\(\dfrac{n_{Fe}}{1}< \dfrac{n_{HCl}}{2}\Rightarrow HCldư\)
Khi đó :
\(n_{Fe}=n_{FeCl_2}=n_{H_2}=0.01\left(mol\right)\)
\(V_{H_2}=0.01\cdot22.4=0.224\left(l\right)\)
\(b.\)
\(n_{HCl\left(dư\right)}=0.1-0.02=0.08\left(mol\right)\)
\(m_{HCl}=0.08\cdot36.5=2.92\left(\right)\)
\(c.\)
\(C_{M_{FeCl_2}}=\dfrac{0.01}{0.1}=0.1\left(M\right)\)
\(C_{M_{HCl\left(dư\right)}}=\dfrac{0.08}{0.1}=0.8\left(M\right)\)
\(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right);nHCl=0,1.1=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,1}{2}\\ \rightarrow HCldư\\ n_{H_2}=n_{Fe}=n_{FeCl_2}=0,01\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\\ V_{ddsau}=V_{ddHCl}=0,1\left(l\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\\ n_{HCl\left(dư\right)}=0,1-0,01.2=0,08\left(mol\right)\\ m_{HCl\left(dư\right)}=0,08.36,5=2,92\left(g\right)\\ C_{MddHCl\left(Dư\right)}=\dfrac{0,08}{0,1}=0,8\left(M\right)\)
