\(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
PTHH:
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
x 2x x x
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
y 2y y y
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
z 2z z z
\(ZnCl_2+2NaOH\rightarrow2NaCl+Zn\left(OH\right)_2\downarrow\)
x 2x 2x x
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
y 2y y 2y
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
z 2z z 2z
\(Zn\left(OH\right)_2\rightarrow\left(t^o\right)ZnO+H_2O\)
x x x
\(4Fe\left(OH\right)_2+O_2\rightarrow\left(t^o\right)2Fe_2O_3+4H_2O\)
y 0,5y y
\(Mg\left(OH\right)_2\rightarrow\left(t^o\right)MgO+H_2O\)
z z z
Ta có Hệ PT:
\(\left\{{}\begin{matrix}x+y+z=0,5\\65x+56y+24z=20,9\\81x+80y+40z=22\end{matrix}\right.\)
Bạn check lại đề có sai ở đâu k nha chứ bậm nghiệm k ra.
Ta có: 65nZn + 56nFe + 24nMg = 20,9 (1)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Zn}+n_{Fe}+n_{Mg}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\left(2\right)\)
PT: \(ZnCl_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaCl\)
\(Zn\left(OH\right)_2+2NaOH\rightarrow Na_2ZnO_2+2H_2O\)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_{2\downarrow}+2NaCl\)
\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_{2\downarrow}+2NaCl\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{FeCl_2}=\dfrac{1}{2}n_{Fe}\\n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}=n_{Mg}\end{matrix}\right.\)
\(\Rightarrow m_{ran}=m_{Fe_2O_3}+m_{MgO}=160.\dfrac{1}{2}n_{Fe}+40n_{Mg}=22\left(g\right)\left(3\right)\)
Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\\n_{Mg}=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Fe}=0,15.56=8,4\left(g\right)\\m_{Mg}=0,25.24=6\left(g\right)\end{matrix}\right.\)