a)
Gọi số mol Mg, FeO là a, b (mol)
=> 24a + 72b = 13,68 (1)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a----->2a------->a------>a
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
b------>2b------->b------>b
=> 95a + 127b = 30,45 (2)
(1)(2) => a = 0,12 (mol); b = 0,15 (mol)
=> \(\left\{{}\begin{matrix}m_{Mg}=0,12.24=2,88\left(g\right)\\m_{FeO}=0,15.72=10,8\left(g\right)\end{matrix}\right.\)
b) nHCl = 2a + 2b = 0,54 (mol)
=> mHCl = 0,54.36,5 = 19,71 (g)
=> \(m_{dd.HCl}=\dfrac{19,71.100}{14,6}=135\left(g\right)\)
mY = 135 + 13,68 - 0,12.2 = 148,44 (g)
c) \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,12.95}{148,44}.100\%=7,68\%\\C\%_{FeCl_2}=\dfrac{0,15.127}{148,44}.100\%=12,83\%\end{matrix}\right.\)
a)
Gọi $n_{Mg} =a (mol) ; n_{FeO} = b(mol)\Rightarrow 24a + 72b = 13,68(1)$
$Mg + 2HCl \to MgCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$
$m_{muối} = m_{MgCl_2} + m_{FeCl_2} = 95a + 127b = 30,45(2)$
Từ (1)(2) suy ra a = 0,12 ; b = 0,15
$m_{Mg} = 0,12.24 = 2,88(gam)$
$m_{FeO} = 0,15.72 = 10,8(gam)$
b)
$n_{HCl} = 2n_{Mg} + 2n_{FeO} = 0,54(mol)$
$m_{dd\ HCl} = \dfrac{0,54.36,5}{14,6\%} = 135(gam)$
$m_{dd\ Y} = m_X + m_{dd\ HCl} - m_{H_2} = 13,68 + 135 - 0,12.2 = 148,44(gam)$
c)
$C\%_{MgCl_2} = \dfrac{0,12.95}{148,44}.100\% = 7,67\%$
$C\%_{FeCl_2} = \dfrac{0,15.127}{148,44}.100\% = 12,83\%$
