\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{HCl}=\dfrac{300\cdot3.65\%}{36.5}=0.3\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Lập tỉ lệ :
\(\dfrac{0.1}{1}< \dfrac{0.3}{2}\) => HCl dư
\(n_{FeCl_2}=n_{H_2}=0.1\left(mol\right)\)
\(n_{HCl\left(dư\right)}=0.3-0.1\cdot2=0.1\left(mol\right)\)
\(m_{dd}=5.6+300-0.1\cdot2=305.4\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{0.1\cdot127}{305.4}\cdot100\%=4.15\%\)
\(C\%_{HCl\left(dư\right)}=\dfrac{0.1\cdot36.5}{305.4}\cdot100\%=1.2\%\)
n(Fe)=0,1mol
m(HCl)=10,95g
n(Hcl)=0,3mol
Fe+2HCL-> FeCl2+H2
Ta có 0,1/1 < 0,3/2
Hcl dư
Fe+2HCl->FeCl2+H2
0,1 0,2 0,1 0,1
mdd =300+5,6-0,1.2=305,4g
C%(Fecl2)=0,1.127.100%/305,4 =4,16%
