a. \(n_{HCl}=1,25.0,4=0,5\left(mol\right)\)
\(n_{H_2}=\dfrac{8.96}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow HCl.dư\)
b. \(n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\)
\(\Rightarrow24a+27b=7,8\left(g\right)\left(1\right)\)
Mg + 2HCl -> MgCl2 + H2
2Al + 6HCl -> 2AlCl3 + 3H2
\(\Rightarrow1,5a+b=0,4\left(mol\right)\left(2\right)\)
Từ ( 1 ) , (2) => a = 0,2 ; b = 0,1
b. \(C_{M_{AlCl_3}}=\dfrac{0.2}{8,96}=0,02\left(M\right)\)
\(C_{M_{MgCl_2}}=\dfrac{0.1}{8,96}=0,01\left(M\right)\)