Đáp án:
$V=538,36ml\ hoặc\ V=470,66ml$
Giải thích các bước giải:
$n_{Al}=\dfrac{10,8}{27}=0,4; n_{H_2}=0,48\ mol$
$2yAl+3Fe_xO_y\xrightarrow{t^o}yAl_2O_3+3xFe$
Gọi $n_{Al\ pư}=a⇒n_{Al\ dư}=0,4-a; n_{Fe}=\dfrac{3x}{2y}.a$
$2Al+6HCl\to 2AlCl_3+3H_2\\Fe+2HCl\to FeCl_2+H_2$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$
$Fe_xO_y+2yHCl\to xFeCl_{\frac{2y}{x}}+yH_2O$
$⇒n_{H_2}=\dfrac{3}{2}.(0,4-a)+\dfrac{3x}{2y}.a ⇒a(1-\dfrac{x}{y})=0,08$
$⇒\dfrac{x}{y}<1⇒\dfrac{x}{y}=\dfrac{2}{3}\ hoặc\ \dfrac{x}{y}=\dfrac{3}{4}$
TH1: $\dfrac{x}{y}=\dfrac{2}{3}⇒a=0,24;\ n_{Fe_2O_3}=0,2175\ mol$
DD sau phản ứng chứa $Al_2(SO_4)_3; Fe_2(SO_4)_3$
BT Al, Fe: $n_{Al_2(SO_4)_3}=0,2; n_{Fe_2(SO_4)_3}=0,2175$
BT S ⇒ $n_{H_2SO_4}=3.n_{Al_2(SO_4)_3}+3. n_{Fe_2(SO_4)_3}=1,2525\ mol$
$⇒m_{H_2SO_4}=122,745g ⇒m_{dd\ H_2SO_4}=613,725g⇒V=538,36ml$
TH2: $\dfrac{x}{y}=\dfrac{3}{4}⇒a=0,32; $n_{Fe_3O_4}=0,15\ mol$
DD sau phản ứng chứa $Al_2(SO_4)_3; Fe_2(SO_4)_3;FeSO_4$
BT Al, Fe: $n_{Al_2(SO_4)_3}=0,2$
Theo PTHH; $n_{FeSO_4}=0,36\ mol$
BT Fe: $n_{Fe_2(SO_4)_3}= \dfrac{0,15.3-0,36}{2}=0,045\ mol$
BT S ⇒ $n_{H_2SO_4}=3.n_{Al_2(SO_4)_3}+3. n_{Fe_2(SO_4)_3}+n_{FeSO_4}=1,095\ mol$
$⇒m_{H_2SO_4}=107,31g ⇒m_{dd\ H_2SO_4}=536,55g⇒V=470,66ml$
