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\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 16,8 - 6,4 = 10,4(1)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)\)
Từ (1)(2) suy ra: a = 0,1 ; b = 0,2
Vậy :
\(\%m_{Fe} = \dfrac{0,1.56}{16,8}.100\% = 33,33\%\\ \%m_{Mg} = \dfrac{0,2.24}{16,8}.100\% = 28,57\%\\ \%m_{Cu} = 100\% - 33,33\% - 28,57\% = 38,1\%\)
\(n_{Cu} = a ; n_{Al} = b ; n_{Fe} = c(mol)\\ \Rightarrow 64a + 27b + 56c = 28,6(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\\ \text{Mặt khác} : n_{O_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3\\ \)
Ta có :
\(\dfrac{n_X}{n_{O_2}}=\dfrac{a+b+c}{0,5a +0,75b + 0,75c} = \dfrac{0,6}{0,4}(3)\\ (1)(2)(3)\Rightarrow a = \dfrac{317}{1460} ; b = \dfrac{121}{365}; c = \dfrac{15}{146}\\ \%m_{Cu} = \dfrac{\dfrac{317}{1460}.64}{28,6}.100\% = 48,59\%\\ \%m_{Al} = \dfrac{\dfrac{121}{365}.27}{28,6}.100\% = 31,3\%\\ \%m_{Fe} = 100\% - 41,59\% - 31,3\% = 27,11\%\)
Gọi số mol của Cu, Fe, Al trong 23,8 gam hhX lần lượt là x, y, z mol
→ mX = 64x + 56y + 27z = 23,8 (1)
\(n_{Cl_2}\) = x + 1,5y + 1,5z = 0,65 (2)
0,25 mol X + HCl → 0,2 mol H2 nên 0,2.(x + y + z) = 0,25.(y + 1,5z) (3)
Từ (1), (2), (3) => x = 0,2 mol; y = 0,1 mol; z = 0,2 mol
\(\%_{Cu} = \dfrac{0,2. 64}{23,8} \approx 53,78\%\)
\(\%_{Fe} = \dfrac{0,1 .56}{23,8} \approx 23,53\%\)
%Al ≈ 22,69%
\(n_{Cu}=a\left(mol\right),n_{Fe}=b\left(mol\right),n_{Al}=c\left(mol\right)\)
\(m_X=64a+56b+27b=35.7\left(g\right)\left(1\right)\)
\(n_{Cl_2}=\dfrac{21.84}{22.4}=0.975\left(mol\right)\)
\(Cu+Cl_2\underrightarrow{^{^{t^0}}}CuCl_2\)
\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}FeCl_3\)
\(Al+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}AlCl_3\)
\(n_{Cl_2}=a+1.5b+1.5c=0.975\left(mol\right)\left(2\right)\)
\(n_{hh}=ka+kb+kc=0.25\left(mol\right)\)
\(n_{H_2}=kb+k\cdot1.5c=0.2\left(mol\right)\)
\(\Leftrightarrow a-0.25b-0.875c=0\left(3\right)\)
\(\left(1\right),\left(2\right),\left(3\right):a=0.3,b=0.15,c=0.3\)
\(\%Cu=\dfrac{0.3\cdot64}{35.7}\cdot100\%=53.78\%\)
\(\%Fe=\dfrac{0.15\cdot56}{35.7}\cdot100\%=23.52\%\)
\(\text{%Al=22.7%}\)
