Question

Answer 1

a)

Mg + 2HCl --> MgCl₂ + H₂

2Al + 6HCl --> 2AlCl₃ + 3H₂

Gọi số mol Mg, Al là a, b (mol)

=> 24a + 27b = 9(1)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

Theo PTHH: \(n_{H_2}=a+1,5b=0,45\) (2)

(1)(2) => a = 0,15 (mol); b = 0,2 (mol)

\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{9}.100\%=40\%\\\%m_{Al}=\dfrac{0,2.27}{9}.100\%=60\%\end{matrix}\right.\)

b) Theo PTHH: \(n_{HCl}=2.n_{H_2}=0,9\left(mol\right)\)

=> m_HCl = 0,9.36,5 = 32,85 (g)

=> \(a\%=\dfrac{32,85}{400}.100\%=8,2125\%\)