Question

Giúp em với please:'((((

Answer 1

a) Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) => 24x + 27y = 2,88 (1)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 

Mg + 2HCl ---> MgCl₂ + H₂ 

x------------------>x------->x

2Al + 6HCl ---> 2AlCl₃ + 3H₂ 

y------------------>y-------->1,5y

=> x + 1,5y = 0,15 (2)

Từ (1), (2) => x = 0,03; y = 0,08

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,03.24}{2,88}.100\%=25\%\\\%m_{Al}=100\%-25\%=75\%\end{matrix}\right.\)

b) m_{dd sau phản ứng} = 2,88 + 397,42 - 0,15.2 = 400 (g)

=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,03.95}{400}.100\%=0,7125\%\\C\%_{AlCl_3}=\dfrac{0,08.133,5}{400}.100\%=2,67\%\end{matrix}\right.\)