Đáp án:
Câu 2:
Gọi CTHH cần tìm là $Al_xS_yO_z$
$\to x:y:z=\dfrac{\%m_{Al}}{27}:\dfrac{\%m_S}{32}:\dfrac{\%m_O}{16}=0,55:0,87:4,9375=2:3:12$
$\to CTDGN: Al_2S_3O_{12}$ hay $Al_2{(SO_4)}_3$
$\to M_{Al_2{(SO_4)}_3}=342$
$\to 342n=342$
$\to n=2$
$\to Al_2{(SO_4)}_3$
Câu 3:
$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$
$a)$ $Fe+2HCl\to FeCl_2+H_2$
Theo PTHH $\to n_{H_2}=n_{Fe}=0,2(mol)$
$\to V_{H_2}=0,2.22,4=4,48(l)$
$b)$ \[n_{HCl}=2n_{Fe}=2.0,2=0,4(mol)\]
\[\to m_{HCl}=36,5.0,4=14,6(g)\]
$c)$ $n_{FeCl_2}=n_{Fe}=0,2(mol)$
$\to m_{FeCl_2}=0,2.127=25,4(g)$
