Giải thích các bước giải:
a,
Gọi số mol Fe, Al lần lượt là x, y
$\to 56x+27y=2,2$
$n_S=\dfrac{2,56}{32}=0,08\ (mol)$
$Fe+S\xrightarrow{t^\circ} FeS$
$2Al+3S\xrightarrow{t^\circ} Al_2S_3$
$\to x+1,5y = 0,08$
$\to \begin{cases} x=0,02\\ y=0,04\end{cases}$
$\to m_{Fe}=0,02.56=1,12\ (gam)$
$\to\%m_{Fe}=\dfrac{1,12}{2,2}\cdot 100\%=50,9\%$
$\to\%m_{Al}=100\%-50,9\%=49,1\%$
b,
BT S: $n_{FeS}=n_{Fe}=0,02\ (mol)$
BT AL: $n_{Al_2S_3}=0,02\ (mol)$
$FeS+2HCl\to FeCl_2+H_2S$
$Al_2S_3+6HCl\to 2AlCl_3+3H_2S$
$\to n_{HCl}=0,02.2+0,02.6=0,16 \ (mol)$
$\to m_{dd\ HCl}=\dfrac{0,16.36,5}{10\%}=58,4\ (gam)$
BT S: $n_{H_2S}=0,02+0,02.3=0,08\ (mol)$
$\to V=0,08.22,4=1,792$ (lít)
BTKL: $m_{dd}=2,2+2,56+58,4-0,08.34=60,44\ (gam)$
$\to C\%_{FeCl_2}=\dfrac{0,02.127}{60,44}\cdot 100\%=4,2\%$
$C\%_{AlCl_3}=\dfrac{0,02.2.133,5}{60,44}\cdot 100\%=8,835\%$
c,
$d_{X/N_2}=\dfrac{34}{28}=1,214$
