nHCl = \(\dfrac{91,25.20\%}{36,5}=0,5\left(mol\right)\)
Gọi: \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Fe_2O_3}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
a 2a a a
Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
b 6b 2b 3b
Hệ phương trình: \(\left\{{}\begin{matrix}56a+160b=13,6\\2a+6b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Fe_2O_3}=0,05.160=8\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{13,6}=41,17\%\\\%m_{Fe_2O_3}=100\%-41,17\%=58,83\%\end{matrix}\right.\)
mH2 = 0,1 . 2 = 0,2 (g)
=> \(m_{dd}=91,25+13,6-0,2=104,65\left(g\right)\)
nFeCl3 = 0,1 + 0,05.2 = 0,2 (mol)
=> mFeCl3 = 0,2.162,5 = 32,5 (g)
=> \(C\%_{FeCl_3}=\dfrac{32,5}{104,65}=31,05\%\)
PTHH:
Fe + 2H2SO4 ---> Fe2(SO4)3 + SO2 + 2H2O
0,1 0,1
Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O
mddNaOH = 64.1,025 = 65,6 (g)
=> nNaOH = \(\dfrac{65,6.10\%}{40}=0,164\left(mol\right)\)
T = \(\dfrac{0,164}{0,1}=1,64\) => phản ứng tạo cả 2 muối
Gọi \(\left\{{}\begin{matrix}n_{NaOH\left(tạo.muối.axit\right)}=x\left(mol\right)\\n_{NaOH\left(tao.nuôi.trung.hoà\right)}=y\left(mol\right)\end{matrix}\right.\)
PTHH:
NaOH + SO2 ---> NaHSO3
x x x
2NaOH + SO2 ---> Na2SO3 + H2O
y \(\dfrac{y}{2}\) y
Hệ phương trình: \(\left\{{}\begin{matrix}x+y=0,164\\x+\dfrac{y}{2}=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,036\left(mol\right)\\y=0,128\left(mol\right)\end{matrix}\right.\)
Đổi 64ml = 0,064l
\(\Rightarrow\left\{{}\begin{matrix}C_{MNaHCO_3}=\dfrac{0,036}{0,064}=0,5625M\\C_{MNa_2CO_3}=\dfrac{0,128}{0,064}=2M\end{matrix}\right.\)
