1.
$n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$
$n_O=\dfrac{2,4}{16}=0,15(mol)$
Đặt CTHH hợp chất là $Fe_2O_x$
$\to \dfrac{2}{x}=\dfrac{0,1}{0,15}=\dfrac{2}{3}$
$\to x=3$
Vậy CTHH hợp chất là $Fe_2O_3$
2.
$2K+2H_2O\to 2KOH+H_2$
$n_K=\dfrac{3,9}{39}=0,1(mol)$
Theo PTHH:
$n_{KOH}=n_K=0,1(mol)$
a,
$m_{H_2O}=DV=200g$
$n_{H_2}=\dfrac{n_K}{2}=0,05(mol)$
$\to m_{dd\rm spu}=3,9+200-0,05.2=203,8g$
$\to C\%_{KOH}=\dfrac{0,1.56.100}{203,8}=2,75\%$
b,
$V_{dd\rm spu}=200ml=0,2l$
$\to C_{M_{KOH}}=\dfrac{0,1}{0,2}=0,5M$