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\(a)n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}1,5a+b=0,2\\27a+56b=5,5\end{matrix}\right.\\ a=0,1\\ b=0,05\\ \%_{Al}=\dfrac{0,1.27}{5,5}\cdot100=49\%\\ \%_{Fe}=100-49=51\%\\ b)n_{HCl\left(1\right)_{ }}=0,1\cdot\dfrac{6}{2}=0,3\left(mol\right)\\ n_{HCl\left(2\right)}=0,05.2=0,1\left(mol\right)\\ n_{HCl}=0,3+0,1=0,4\left(mol\right)\\ C_{M_{HCl}}=\dfrac{0.4}{0,5}=0,8M\)
Sửa đề : 13.9 (g)
\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)
\(\Rightarrow m=27a+56b=13.9\left(1\right)\)
\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=1.5a+b=0.35\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(\%Al=\dfrac{0.1\cdot27}{13.9}\cdot100\%=19.42\%\)
\(\%Fe=100-19.42=80.58\%\)
2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2 (1)
Fe + 2HCl \(\rightarrow\)FeCl2 + H2 (2)
nH2=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Đặt nAl=a\(\Leftrightarrow m_{Al}=27a\)
nFe=b\(\Leftrightarrow m_{Fe}=56b\)
Ta có hệ pt:
\(\left\{{}\begin{matrix}27a+56b=5,5\\\dfrac{3}{2}a+b=0,2\end{matrix}\right.\)
Giải hệ ta có:
a=0,1;b=0,05
mAl=27.0,1=2,7(g)
C% Al=\(\dfrac{2,7}{5,5}.100\%=49\%\)
C% Fe=100-49=51%
b;\(\sum\)nHCl=0,1.3+0,05.2=0,4(mol)
CMHCl=\(\dfrac{0,4}{0,5}=0,8M\)
nH2 = 2,352 = 0,098 (mol)
mH2 = 0,098 . 2 = 0,196 (g)
PTHH:
Mg + 2HCl -> MgCl2 + H2
2Al + 6HCl -> 2AlCl3 + 3H2
Fe + 2HCl -> FeCl2 + H2
Từ PTHH: nHCl = 2.nH2 = 2 . 0,098 = 0,196 (mol)
mHCl = 0,196 . 36,5 = 7,154 (g)
Áp dụng ĐLBTKL, ta có:
mkl + mHCl = mmuối + mH2
=> mmuối = 3,53 + 7,154 - 0,196 = 10,448 (g)
Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 56y = 5,5 (1)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,5}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)
\(n_{O_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(a.......\dfrac{2a}{3}\)
\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)
\(b.......\dfrac{3b}{4}\)
\(n_{O_2}=\dfrac{2a}{3}+\dfrac{3b}{4}=0.25\left(mol\right)\left(1\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{H_2}=a+1.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.15,b=0.2\)
\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)
\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)
\(\%m_{Fe}=\dfrac{8.4}{8.4+5.4}\cdot100\%=60.8\%\)
\(\%m_{Al}=100-60.8=39.2\%\)
đề hỏi gì vậy bạn?
đề hỏi tại sao Thắng lại đẹp zaiii qa dọ :3