Bạn tham khảo!

Ta có 2 PTHH sau:

$\rightarrow$ $\text{1) 4Al+3O2$\xrightarrow{t^o}$2Al2O3}$

$\rightarrow$ $\text{2) Al2O3+3H2SO4$\rightarrow$Al2(SO4)3+3H2O}$

$\text{nO2=$\frac{V}{22,4}$=$\frac{3,36}{22,4}$=0,15 mol}$

Theo PTHH 1) $\text{nAl=$\frac{0,15.4}{3}$=0,2 mol}$

$\rightarrow$ $\text{a) mAl=n.M=0,2.27=5,4g}$

Theo PTHH 1) $\text{nAl2O3=$\frac{0,15.2}{3}$=0,1 mol}$

$\rightarrow$ $\text{b) mAl2O3=n.M=0,1.102=10,2g}$

Theo PTHH 2) $\text{nH2SO4=3nAl2O3=0,1.3=0,3 mol}$

$\rightarrow$ $\text{c) V=$\frac{n}{CM}$=$\frac{0,3}{0,5}$=0,6M}$

$\textit{Bạn tham khảo nha:}$

$\textit{PTHH: 4 AL + 3 $O_2$ ->2$AL_2$$O_3$(điều kiện nhiệt)}$

$\textit{n$O_2$=$\frac{3,36}{22,4}$=0,15(mol)}$

$\textit{Dựa vào phương trình =>nAL= $\frac{4}{3}$ *0,15=0,2(mol)}$

$\textit{=>mAL đã tham gia phản ứng = 0,2*27=5,4(g)}$

$\textit{$AL_2$$O_3$=0,5nAl=0,5*0,2=0,1(mol)}$

$\textit{=>m$AL_2$$O_3$=0,1*(27*2+16*3)=10,2(g)}$

$\textit{c, PTHH : $AL_2$$O_3$ + 3$H_2$S$O_4$ -> $Al_2(SO_4)_3$ }$ + 3$H_2$O

$\textit{Dựa vào phương trình =>n$H_2$S$O_4$=3n$AL_2$$O_3$=3*0,1=0,3(mol)}$

$\textit{=>V $H_2$S$O_4$ cần dùng =$\frac{0,3}{0,5}$ =0,6(M)}$