Fe+2HCl->FeCl2+H2
x---------------------x
FeS+2HCl->FeCl2+H2S
y-------------------------y
gọi số mol của fe,fes lần lượt là x,y
->ta có hệ 56x+88y=19,2 và x+y=0,24 -> x=0,06 y=0,18
-> mfe=0,06*56=3,36g , mfes =0,18*88=15,84g
=>mHCl=0,24.36,5=8,76g
=>mddHCl=43,8g
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
Gọi số mol Fe là x; FeS là y.
\(\Rightarrow56x+88y=19,2\)
\(n_H+n_{H2S}=n_{Fe}+n_{FeS}=x+y=\frac{5,376}{22,4}=0,24\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,18\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,06.56=3,36\left(l\right)\)
\(\Rightarrow m_{FeS}=19,2-3,36=15,84\left(g\right)\)
\(n_{HCl}=2n_{H2}+2n_{H2S}=0,24.2=0,48\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,48.36,5=17,52\left(g\right)\)
\(\Rightarrow m_{dd\left(HCl\right)}=\frac{17,52}{20\%}=87,6\left(g\right)\)
BTKL:
\(m_{dd\left(spu\right)}=19,2+87,5-m_{H2}-m_{H2S}=19,2+87,5-0,06.2-0,18.34=100,56\left(g\right)\)
\(n_{FeCl2}=n_{Fe}+n_{FeCl2}=0,24\left(mol\right)\)
\(\Rightarrow m_{FeCl2}=0,25.\left(56+35,5.2\right)=30,48\left(g\right)\)
\(\Rightarrow\%m_{FeCl2}=\frac{30,48}{100,56}=30,3\%\)