a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(m_{HCl}=200.29,2\%=58,4\left(g\right)\Rightarrow n_{HCl}=\dfrac{58,4}{36,5}=1,6\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) => 24a + 56b = 10,4 (*)
PTHH
Mg + 2HCl ---> MgCl2 + H2 (1)
a----->2a-------->a-------->a
Fe + 2HCl ---> FeCl2 + H2 (2)
b---->2b------->b-------->b
Theo (1), (2): nHCl (phản ứng) = 2nH2 = 0,6 (mol) < 1,6 (mol)
=> HCl dư
b) a + b = 0,3 (**)
Từ (*), (**) => a = 0,2; b = 0,1
nHCl dư = 1,6 - 0,6 = 1 (mol)
mdd sau phản ứng = 10,4 + 200 - 0,3.2 = 209,8 (g)
=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{209,8}.100\%=9,06\%\\C\%_{FeCl_2}=\dfrac{0,1.127}{209,8}.100\%=6,05\%\\C\%_{HCl.dư}=\dfrac{1.36,5}{209,8}.100\%=17,4\%\end{matrix}\right.\)
a) $n_{HCl} = \dfrac{200.29,2\%}{36,5} = 1,6(mol)$
$n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH, $n_{HCl\ pư} = 2n_{H_2} = 0,6 < 1,6$
$\Rightarrow$ axit dư
b) Gọi $n_{Mg} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 24a + 56b = 10,4(1)$
Theo PTHH, $n_{H_2} = a + b = 0,3(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,1
$m_{dd\ sau\ pư} = 10,4 + 200 - 0,3.2 = 109,8(gam)$
$C\%_{MgCl_2} = \dfrac{0,2.95}{109,8}.100\% = 17,3\%$
$C\%_{FeCl_2} = \dfrac{0,1.127}{109,8}.100\% = 11,57\%$