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)