a,
$n_{H_2SO_4}=\dfrac{375.23,52\%}{98}=0,9(mol)$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$Ba+H_2SO_4\to BaSO_4+H_2$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$\to n_{H_2SO_4\rm pứ}=n_{H_2}=0,3(mol)<0,9$
$\to H_2SO_4$ dư, kim loại tan hết trong axit
Đặt $x$, $y$ là số mol $Ba$, $Al$
$\to 137x+27y=14,325$
Ta có: $n_{Ba}+1,5n_{Al}=n_{H_2}$
$\to x+1,5y=0,3$
Giải hệ ta có: $x=0,075; y=0,15$
$m_{Ba}=0,075.137=10,275g$
$m_{Al}=0,15.27=4,05g$
b,
$n_{BaSO_4}=0,075(mol)$
$\to m_A=14,325+375-0,075.233-0,3.2=371,25g$
$n_{H_2SO_4\rm dư}=0,9-0,3=0,6(mol)$
$\to C\%_{H_2SO_4}=\dfrac{0,6.98.100}{371,25}=15,83\%$
$n_{Al_2(SO_4)_3}=0,5y=0,075(mol)$
$\to C\%_{Al_2(SO_4)_3}=\dfrac{0,075.342.100}{371,25}=6,9\%$