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\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)
\(m=80a+160b=20\left(g\right)\left(1\right)\)
\(n_{HCl}=0.2\cdot3.5=0.7\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(n_{HCl}=2a+6b=0.7\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.05,b=0.1\)
\(m_{CuO}=0.05\cdot80=4\left(g\right)\)
\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)
ta có phương trình:
2Al+6HCl=>2AlCl3+3H2
a 1.5a
Mg+2HCl=>MgCl2+H2
b b
ta có vH2=13.44(lít)=>nH2=13.4422.4=0.6(mol)
gọi a là số mol của Al,b là số mol của Mg
=>1.5a+b=0.6(mol)(1)
27a+24b=12.6(g)(2)
từ (1)(2)=>a=0.2(mol),b=0.3(mol)
=>%Al=0.2∗2712.6*100=42.86%
=>%Mg=100-42.86=57.14%
Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Mg}=y\end{matrix}\right.\) ( mol ) \(\rightarrow m_{hh}=27x+24y=12,6\left(g\right)\) (1)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x 3x 1,5x ( mol )
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
y 2y y ( mol )
\(n_{H_2}=1,5x+y=\dfrac{13,44}{22,4}=0,6\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{12,6}.100=42,85\%\\\%m_{Mg}=100-42,85=57,15\%\end{matrix}\right.\)
\(n_{HCl}=3.0,2+2.0,3=1,2\left(mol\right)\)
\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(a.2Al+6HCl->2AlCl_3+3H_2\\ Mg+2HCl->MgCl_2+H_2\\ b.n_{Al}=a,n_{Mg}=b\\ 27a+24b=7,5\left(I\right)\\ 1,5a+b=\dfrac{7,84}{22,4}=0,35\left(II\right)\\ a=0,1;b=0,2\\ \%m_{Al}=\dfrac{27\cdot0,1}{7,5}\cdot100\%=36\%\\ \%m_{Mg}=64\%\\ c.m_{HCl}=36,5\left(0,1\cdot3+0,2\cdot2\right)=18,25g\\ d.m_{ddsau}=7,5+\dfrac{18,25}{14,6:100}-0,35\cdot2=131,8g\\ C\%\left(AlCl_3\right)=\dfrac{133,5\cdot0,1}{131,8}\cdot100\%=10,1\%\\ C\%\left(MgCl_2\right)=\dfrac{95\cdot0,2}{131,8}\cdot100\%=14,4\%\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{18,48}{22,4}=0,825\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}24x+27y=17,1\\x+\dfrac{3}{2}y=0.825\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,3\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,375.24}{17,1}.100=52,63\%\\ \%m_{Al}=47,37\%\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,2
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(m_{Mg}=0,2.24=4,8g\\ m_{MgO}=18,4-4,8=13,6g\)
Các PTHH mà