a.
Gọi $n_{Fe}=x(mol)$
$n_{FeO}=y(mol)$
$n_{FeCO_3}=x(mol)$
$⇒56x+72y+116z=17,28(g)(1)$
$Fe+2HCl→FeCl_2+H_2↑$
x → 2x x x (mol)
$FeO+2HCl→FeCl_2+H_2O$
y → 2y y (mol)
$FeCO_3+2HCl→FeCl_2+CO_2↑+H_2O$
z → 2z z z (mol)
$⇒m_{muối}=127x+127y+127z=25,4(g)$
$⇔x+y+z=0,2 (2)$
-Ta có:$\frac{M_{H_2}.x+M_{CO_2}.y}{x+y}=\frac{2x+44x}{x+y}=15.2=30$
$⇔-28x+14z=0$
$⇔-2x+z=0(3)$
-Từ (1), (2) và (3) ,ta có hệ pt:
\begin{cases}56x+72y+116z=17,28 \\x+y+z=0,2 \\ -2x+z=0\end{cases} \begin{cases} x=0,04\\ y=0,08\\z=0,08 \end{cases}
$-n_{HCl}=0,04.2+0,08.2+0,08.2=0,4(mol)$
$V_{dd...HCl}=\frac{0,4}{0,5}=0,8(l)$
b.
$m_{Fe}=0,04.56=2,24(g)$
⇒%$m_{Fe}=$$\frac{2,24}{17,28}.100$%$≈ 12,962$%
$m_{FeO}=0,08.72=5,76(g)$
⇒%$m_{FeO}=$$\frac{5,76}{17,28}.100$%$≈ 33,33$%
%$m_{FeCO_3}=100$%$-$%$m_{Fe}-$%$m_{FeO}=100$%$-12,962$%$-33,33$%$=53,708$%
