giúp em với :<<<<<<<<<<

$n_{SO_2}=\dfrac{6,72}{22,4}=0,3(mol)$

Gọi chung hh $X$ là $Fe_xO_y$.

$2Fe_xO_y+(6x-2y)H_2SO_4\to xFe_2(SO_4)_3+(3x-2y)SO_2+(6x-2y)H_2O$

$n_{Fe_xO_y}=\dfrac{2}{3x-2y}n_{SO_2}=\dfrac{0,6}{3x-2y}(mol)$

$\to M=\dfrac{75,2(3x-2y)}{0,6}=\dfrac{376}{3}(3x-2y)=56x+16y$

$\to 320x=\dfrac{800}{3}y$

$\to y=\dfrac{6}{5}x$

Suy ra CTTQ hỗn hợp là $Fe_xO_{\frac{6}{5}x}$

$\%m_{Fe}=\dfrac{56x.100}{56x+16.\dfrac{6}{5}x}=\dfrac{3500}{47}\%$

$\to m=75,2.\dfrac{3500}{47}\%=56g$