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
$n_{Fe}=2/3nSO2=0,04(mol)$
=>$m_{Fe}=0,04.56=2,24(g)$
$n{H2SO4}(l,PT1)=nFe=0,04(mol)$
=>$n_{H2SO4}(PT2)=0,1-0,04=0,06(mol)$
$n_{Fe2O3}=1/3nH2SO4=0,02(mol)$
=>$m_{Fe2O3}=0,02.160=3,2(g)$
=>$m_{hh}=3,2+2,24=5,24(g)$
=>%$m_{Fe}=2,24/5,24.100$%$=42,75$%
=>%$m_{Fe2O3}=100-42,75=27,25$%