a) 2Al+3H2SO4--->Al2(SO4)3+3H2
Ta có
n Al=5,4/27=0,2(mol)
Theo pthh
n H2=3/2n Al=0,3(mol)
V H2=0,3.22,4=6,72(l)
b) n H2SO4=3/2n Al=0,3(mol)
m H2SO4=0,3.98=29,4(g)
m H2SO4 dư=29,4.20/100=5,88(g)
m H2SO4 tham gia=29,4-5,88=23,52(g)
m dd H2SO4=23,52.100/9,8=240(g)
m dd sau pư=5,4+240-0,6=244,8(g)
C%H2SO4 dư=5,88/244,8.100%=2,4%
n Al2(SO4)3=1/2n Al=0,1(mol)
m Al2(SO4)3=0,1.342=34,2(g)
C% Al2(SO4)3=34,2/244,8.100%=13,97%