Đáp án:

Giải thích các bước giải:

$Fe+H2SO4(l)--->FeSO4+H2$

$Fe2O3+3H2SO4-->Fe2(SO4)3+3H2O$

$2Fe+6H2SO4--->Fe2(SO4)3+3SO2+6H2O$

$n_{H2SO4(l)}=100.9,8/100.98=0,1(mol)$

$n_{SO2}=1,34/22,4=0,06(mol)$

Theo PTHH3

$n_{Fe}=2/3nSO2=0,04(mol)$

=>$m_{Fe}=0,04.56=2,24(g)$

$n{H2SO4}(l,PT1)=nFe=0,04(mol)$

=>$n_{H2SO4}(PT2)=0,1-0,04=0,06(mol)$

$n_{Fe2O3}=1/3nH2SO4=0,02(mol)$

=>$m_{Fe2O3}=0,02.160=3,2(g)$

=>$m_{hh}=3,2+2,24=5,24(g)$

=>%$m_{Fe}=2,24/5,24.100$%$=42,75$%

=>%$m_{Fe2O3}=100-42,75=27,25$%

Gọi a, b là mol Fe, Fe2O3

$n_{H_2SO_4}= \frac{100.9,8\%}{98}= 0,1 mol$

$Fe+ H_2SO_4 \to FeSO_4+ H_2$

$Fe_2O_3+ 3H_2SO_4 \to Fe_2(SO_4)_3+ 3H_2O$ 

=> $a+3b= 0,1$    (1) 

$n_{SO_2}= 0,06 mol$ 

Bảo toàn e: $3a= 0,06$       (2)

(1)(2) => $a=0,02; b= \frac{2}{75}$

$m_{Fe}= 0,02.56= 1,12g$

$m_{Fe_2O_3}= 160.\frac{2}{75}= 4,267g$

=> $\%Fe= 20,79\%; \%Fe_2O_3= 79,21\%$