2Al+3H2SO4=>Al2(SO4)3+3H2

mAl=7.1(g)=>nAl=$\frac{7.1}{27}$ ≈0.26(mol)

=>nH2=$\frac{0.26*3}{2}$=0.39(mol) =>vH2=0.39*22.4=8.736(lít)

ta có nH2SO4=nH2=0.39(mol)=>a=$\frac{0.39}{0.2}$=1.95 M 

ta có:nAl2(SO4)3=$\frac{0.26}{2}$=0.13(mol) =>mAl2(SO4)3=0.13*(27*2+32*3+16*12)=44.46(g)

+)ta có phương trình:Al2(SO4)3+3BaCl2=>2AlCl3+3BaSO4

nBaSO4=0.13*3=0.39(mol)

=>mBaSO4=0.39*(137+32+16*4)=90.87(g)

+)2Al+6H2SO4=>Al2(SO4)3+6H2O+3SO2

mAl=5.4(g)=>nAl=$\frac{5.4}{27}$=0.2(mol) 

=>nSO2=$\frac{0.2*3}{2}$=0.3(mol) 

=>mSO2=0.3*(32+16*2)=19.2(g)

Giải thích các bước giải:

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = 0,26mol\\
 \to {n_{{H_2}}} = 1,5{n_{Al}} = 0,39mol \to {V_{{H_2}}} = 8,84l\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,39mol \to C{M_{{H_2}S{O_4}}} = \dfrac{n}{V} = 1,95M\\
3BaC{l_2} + A{l_2}{(S{O_4})_3} \to 3BaS{O_4} + 2AlC{l_3}\\
{n_{BaS{O_4}}} = 3{n_{A{l_2}{{(S{O_4})}_3}}} = 3 \times \dfrac{1}{2}{n_{Al}} = 0,39mol\\
 \to {m_{BaS{O_4}}} = 90,87g\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{Al}} = 0,2mol \to {n_{S{O_2}}} = 1,5{n_{Al}} = 0,3mol\\
 \to {m_{S{O_2}}} = 19,2g
\end{array}\)