$Al_2O_3+3H_2SO_4→Al_2(SO_4)_3+3H_2O$
a,$nAl=5,1/102=0,05(mol)$
$⇒nH_2SO_4=3nAl_2O_3=0,15(mol)$
$⇒mH_2SO_4=0,15.98=14,7(g)$
$⇒$$C$%$H_2SO_4=14,7.100/200=7,35$%
b,Dung dịch A là dung dịch $Al_2(SO_4)_3$
$nAl_2O_3=nAl_2(SO_4)_3=0,05(mol)$
$⇒mAl_2(SO_4)_3=0,05.345=17,25(g)$
$⇒mddA=mAl_2O_3+mH_2SO_4=5,1+200=205,1(g)$
$⇒$$C$%$Al_2(SO_4)_3=17,25.100/205,1=8,41$%