- Phần 1:
Gọi: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\\n_{Cu}=z\left(mol\right)\end{matrix}\right.\) (trong phần 1)
⇒ 24x + 27y + 64z = 3,48 (1)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PT: \(n_{HCl}=2n_{Mg}+3n_{Al}=2x+3y=0,16\left(2\right)\)
- Phần 2:
Mg, Al, Cu có số mol lần lượt là: kx, ky, kz (mol)
PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Cu}=\dfrac{1}{2}kx+\dfrac{3}{4}ky+\dfrac{1}{2}kz=0,165\left(3\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{H_2}=n_{CuO}=n_{Cu}=kz=0,09\left(4\right)\)
Từ (3) và (4) có: \(\dfrac{kz}{\dfrac{1}{2}kx+\dfrac{3}{4}ky+\dfrac{1}{2}kz}=\dfrac{0,09}{0,165}\Rightarrow\dfrac{z}{\dfrac{1}{2}x+\dfrac{3}{4}y+\dfrac{1}{2}z}=\dfrac{6}{11}\)
⇒ 3x + 4,5y - 8z = 0 (5)
Từ (1), (2) và (5) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,04\left(mol\right)\\z=0,03\left(mol\right)\end{matrix}\right.\)
Thay vào (4) ⇒ k = 3
Vậy: nMg = x + kx = 0,08 (mol) ⇒ mMg = 0,08.24 = 1,92 (g)
nAl = y + ky = 0,16 (mol) ⇒ mAl = 0,16.27 = 4,32 (g)
nCu = z + kz = 0,12 (mol) ⇒ mCu = 0,12.64 = 7,68 (g)