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Đáp án: `x=20g,y=14g`

a)) Sắt `III` oxit là: `Fe_2O_3`

PTHH xảy ra:

`Fe_2O_3+3H_2->2Fe+3H_2O`

b)`n_(H_2)=(5,6)/(22,4)=0,25mol`

`Fe+2HCl->FeCl_2+H_2`

`0,25←......................0,25mol`

Ta có: `n_(Fe)=n_(H_2)=0,25mol`

`y=m_(Fe)=0,25×56=14g`

Theo PTHH ở câu a ta được:

`n_(Fe_2O_3)=n_(Fe)/2=0,125mol`

`m_(Fe_2O_3)=0,125×(56×2+16×3)=20g=x`