b.\(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{29,2}{36,5}=0,8mol\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Mg}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=27x\\m_{Mg}=24y\end{matrix}\right.\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x 3x ( mol )
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
y 2y ( mol )
Ta có:
\(\left\{{}\begin{matrix}27x+24y=7,8\\3x+2y=0,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4g\\m_{Mg}=0,1.24=2,4g\end{matrix}\right.\)
a,b \(n_{HCl}=\dfrac{29.2}{36,5}=0,8\left(mol\right)\)
Gọi nAl = a ( mol ) ; nMg = b(mol)
PTHH : 2Al + 6HCl -> 2AlCl3 + 3H2 ( 1 )
PTHH : Mg + 2HCl -> MgCl2 + H2 ( 2 )
Ta có : 27a + 24b = 7,8
Theo PT ( 1 ) : nHCl = 3 nAl = 3a (mol)
Theo PT (2 ) : nHCl = 2 nMg =2b (mol )
=> 3a + 2b = 0,8
Từ những điều trên ta có \(\left[{}\begin{matrix}27a+24b=7,8\\3a+2b=0,8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\end{matrix}\right.\)
PT (1) : Ta thấy : 0,2 < 0,8 => Al đủ , HCl dư
\(V_{H_2}=\dfrac{\left(0,2.3\right)}{2}.22,4=6,72\left(l\right)\)
PT (2) : Ta thấy : 0,1 < 0,8 => Mg đủ , HCl dư
\(V_{H_2}=\dfrac{0,1.1}{1}.22,4=2,24\left(l\right)\)
