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,

$n_{BaSO_4}=0,075(mol)$

$\to m_A=14,325+375-0,075.233-0,3.2=371,25g$

$n_{H_2SO_4\rm dư}=0,9-0,3=0,6(mol)$

$\to C\%_{H_2SO_4}=\dfrac{0,6.98.100}{371,25}=15,83\%$

$n_{Al_2(SO_4)_3}=0,5y=0,075(mol)$

$\to C\%_{Al_2(SO_4)_3}=\dfrac{0,075.342.100}{371,25}=6,9\%$

Ba+H2SO4→BaSO4+H2

a                                a

2Al+3H2SO4→Al2(SO4)3+3H2

b                                         1,5b

a) Gọi nBa=a, nAl=b

mH2SO4=375.23,52/100=88,2g

nH2SO4=88,2/98=0,9mol

nH2=6,72/22,4=0,3mol

Vì nH2<nH2SO4→Axit dư

137a+27b=14,325

a+1,5b=0,3

→a=0,075

   b=0,15

mBa=0,075.137=10,275g

%mBa=10,275/14,325.100%=71,73%

%mAl=100-71,73=28,27%

b) mdd spứ=14,325+375-0,075.233=371,85g

mAl2(SO4)3=0,15/2.342=25,65g

C%Al2(SO4)3=25,65/371,85.100%=6,9%

mH2SO4 dư=(0,9-0,3).98=58,8g

C%H2SO4 dư=58,8/371,85.100%=15,81%