nAl=8,1/27=0,3mol ;nH2SO4=53,9/98=0,55mol
ta có pt : 2Al+3H2SO4---->Al2(SO4)3+3H2
Trước p/u: 0,3mol 0,55mol
p/u : 0,3mol 0,45mol
Saup/u: 0mol 0,1mol 0,15mol 0,45mol
=>H2SO4 dư
mH2SO4 dư =0,1.98=9,8g
b,mAl2SO43 =0,15.294=44,1g
c, mH2=0,45.2=0,9g
V H2=0,45.22,4=10,08l