Question

Answer 1

Gọi số mol Fe, Mg là a, b (mol)

=> 56a + 24b = 9,2 (1)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

              a--->2a------>a---->a

            Mg + 2HCl --> MgCl₂ + H₂

              b--->2b------>b---->b

=> a + b = 0,25 (2)

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

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{9,2}.100\%=60,87\%\\\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%=39,13\%\end{matrix}\right.\)

n_HCl = 2a + 2b = 0,5 (mol)

=> \(a=\dfrac{0,5}{0,2}=2,5\)

PTHH: FeCl₂ + 2NaOH --> Fe(OH)₂ + 2NaCl

            0,1------>0,2

            MgCl₂ + 2NaOH --> Mg(OH)₂ + 2NaCl

             0,15--->0,3

=> n_NaOH = 0,2 + 0,3 = 0,5 (mol)

=> \(V_{dd.NaOH}=\dfrac{0,5}{2}=0,25\left(l\right)\)

Answer 2

\(\left\{{}\begin{matrix}Fe:x\left(mol\right)\\Mg:y\left(mol\right)\end{matrix}\right.\)

\(n_Y=\dfrac{5,6}{22,4}=0,25mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

x                                     x

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

y                                         y

\(\Rightarrow\left\{{}\begin{matrix}x+y=0,25\\56x+24y=9,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1mol\\y=0,15mol\end{matrix}\right.\)

\(\%m_{Fe}=\dfrac{0,1\cdot56}{9,2}\cdot100\%=60,87\%\)

\(\%m\%m_{Mg}=100\%-60,87\%=39,13\%\)

\(n_{HCl}=2n_{Fe}+2n_{Mg}=2\cdot0,1+2\cdot0,15=0,5mol\)

\(a=C_M=\dfrac{0,5}{0,2}=2,5M\)