a) m\(_{ddHCl}=58,82.1,19=70\left(g\right)\)
\(m_{HCl}=\frac{70,36,5}{100}=25,55\left(g\right)\)
\(n_{HCl}=\frac{25,55}{36,5}=0,7\left(mol\right)\)
Gọi \(n_{Mg},n_{Fe},n_{Zn}\) lần lượt là x,y,z
\(\Rightarrow24x+56y+65z=24,9\left(1\right)\)
\(Mg+2HCl--->MgCl2+H2\)
x------2x(mol)
\(Fe+2HCl---.FeCl2+H2\)
y------2y(mol)
\(Zn+2HCl-->ZnCl2+H2\)
y------2y(mol)
\(\Rightarrow2x+2y+2z=0,7\Leftrightarrow x+y+z=0,35\)(2)
Từ 1
\(\Rightarrow24x+56y+65z=24,9\)
\(\Leftrightarrow24\left(x+y+z\right)+32y+41z=24,9\)
\(\Leftrightarrow x+y+z=1,0375-34y-41z\)
Do y,z \(\ge0\forall y,z\Rightarrow x+y+z\ge1,0375\)
Mà \(x+y+z=0,35\)
=> KL dư
b)\(n_{H2}=\frac{1}{2}n_{HCl}=0,35\left(mol\right)\)
\(V_{H2}=0,35.22,4=7,84\left(l\right)\)
