Question

Answer 1

a) Gọi số mol Al, Zn là a, b (mol)

=> 27a + 65b = 3,14 (1)

\(n_{H_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             a-------------->a------->1,5a

            Zn + 2HCl --> ZnCl₂ + H₂

              b-------------->b----->b

=> 1,5a + b = 0,07 (2)

(1)(2) => a = 0,02 (mol); b = 0,04 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02.27}{3,14}.100\%=17,197\%\\\%m_{Zn}=\dfrac{0,04.65}{3,14}.100\%=82,803\%\end{matrix}\right.\)

b) 

m_{dd sau pư} = 3,14 + 200 - 0,07.2 = 203 (g)

\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,02.133,5}{203}.100\%=1,315\%\\C\%_{ZnCl_2}=\dfrac{0,04.136}{203}.100\%=2,68\%\end{matrix}\right.\)