Đặt $a$, $b$ là số mol $NO$, $NO_2$
$\to a+b=0,25$
$\overline{M}_Y=19.2=38$
$\to 30a+46b=38.0,25=9,5$
Giải hệ ta có: $a=b=0,125$
Gọi chung hh oxit, kim loại là $Fe_xO_y$
$3Fe_xO_y+(12x-2y)HNO_3\to 3xFe(NO_3)_3+(3x-2y)NO+(6x-y)H_2O$
$Fe_xO_y+(6x-2y)HNO_3\to xFe(NO_3)_3+(3x-2y)NO_2+(3x-y)H_2O$
Theo PTHH:
$n_{Fe_xO_y}=\dfrac{0,125.3}{3x-2y}+\dfrac{0,125}{3x-2y}=\dfrac{0,5}{3x-2y}$
$\to M=\dfrac{20(3x-2y)}{0,5}=40(3x-2y)=56x+16y$
$\to 64x=96y$
$\to y=\dfrac{2}{3}x$
Suy ra CTTQ hh là $Fe_xO_{\frac{2}{3}x}$
$\%m_{Fe}=\dfrac{56x.100}{56x+16.\dfrac{2}{3}x}=84\%$
$\to m=20.84\%=16,8g$
Ta có:
$n_{HNO_3}=\dfrac{12x-2y}{3x-2y}n_{NO}+\dfrac{6x-2y}{3x-2y}n_{NO_2}=\dfrac{12x-2.\dfrac{2}{3}x}{3x-2.\dfrac{2}{3}x}.n_{NO}+\dfrac{6x-2.\dfrac{2}{3}x}{3x-2.\dfrac{2}{3}x}n_{NO_2}= 6,4.0,125+2,8.0,125=1,15(mol)$
$\to V_{HNO_3}=1,15l$
