a, PTHH
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_x_____3x_______x______1,5x
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
_y_____2y_______y_______y
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
ta có hệ\(\left\{{}\begin{matrix}27x+24y=6,3\\1,5x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1\times27}{6,3}=42,86\%\\\%m_{Mg}=100\%-42,86\%=57,14\%\end{matrix}\right.\)
b,\(n_{HCl}=3x+2y=0,6\left(mol\right)\)
thể tích dung dịch HCl cần dùng là:
\(V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,6}{0,4}=1,5\left(l\right)=1500\left(ml\right)\)
6,3 gam hôn hợp \(\left\{{}\begin{matrix}Al:a\left(mol\right)\\Mg:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow27a+24b=6,3\)(I)
\(2Al\left(a\right)+6HCl\left(3a\right)--->2AlCl_3\left(a\right)+3H_2\left(1,5a\right)\)
\(Mg\left(b\right)+2HCl\left(2b\right)--->MgCl_2\left(b\right)+H_2\left(b\right)\)
\(n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow1,5a+b=0,3\)(II)
từ (I) và (II): \(\left\{{}\begin{matrix}27a+24b=6,3\\1,5a+b=0,3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\)
=> Phần trăm khối lượng
\(b)\)
Theo PTHH: \(n_{HCl}=3a+2b=3.0,1+2.0,15=0,6\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,4}=1,5\left(l\right)=1500\left(ml\right)\)
\(c)\)
Vì phản ứng vùa đủ
\(\Rightarrow\)Dung dịch sau phản ứng: \(\left\{{}\begin{matrix}AlCl_3:0,1\left(mol\right)\\MgCl_2:0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{H_2}=0,3.2=0,6\left(g\right)\)
\(m_{ddHCl}=D_{HCl}.V_{ddHCl}=1,2.1500=1800\left(g\right)\)
\(\Rightarrow m_{ddsau}=6,3+1800-0,6=1805,7\left(g\right)\)
=> Tính được C% mỗi muối
