Đáp án:
Giải thích các bước giải:
$nAl=10,8:27=0,4$(mol)
$2Al + 3H2SO4 => Al2(SO4)3 + 3H2↑$
a,$nH2=3/2 .nAl=3/2.0,4=0,6$(mol)
$VB=VH2=0,6.22,4=13,44(l)$
b, $nH2SO4=3/2.nAl=3/2.0,4=0,6$(mol)
$=> mH2SO4=0,6.98=58,8(g)$
$=>$%$H2SO4=\frac{58,8}{300}.100=19,6 $%
c, $nAl2(SO4)3=1/2.n=1/2.0,4=0,2$(mol)
$=>mAl2(SO4)3=0,2.342=68,4(g)$
$mdd mới=mAl+mH2SO4-mH2=10,8+300-0,6.2=309,6(g)$
$=>$%$Al2(SO4)3=68,4:309,6.100=22,1$%
