Question

Cho 12,1 (g) h2 Fe,Zn tác dụng HCl dư 4,48(l) H2 đktc . Xđ %m

Answer 1

Gọi số mol Fe, Zn là a, b (mol)

=> 56a + 65b = 12,1 (1)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

             a----------------------->a

            Zn + 2HCl --> ZnCl₂ + H₂

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

=> a + b =0,2 (2)

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

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\\\%m_{Zn}=\dfrac{0,1.65}{12,1}.100\%=53,72\%\end{matrix}\right.\)