Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) => 27a + 56b = 11,1 (*)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ m_{H_2SO_4}=200.19,6\%=39,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0,45.1=0,45\left(mol\right)\)
PTHH:
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (2)
Theo PT (1), (2): \(n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,3\left(mol\right)< 0,4\left(mol\right)=n_{H_2SO_4\left(bđ\right)}\)
=> H2SO4 dư, hh Al, Fe tan hết
Theo PT (1), (2): \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y=0,3\left(mol\right)\) (**)
Từ (*), (**) => x = 0,1; y = 0,15
Theo PT (1): \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\)
Theo PT (2): \(n_{FeSO_4}=n_{Fe}=0,15\left(mol\right)\)
\(n_{H_2SO_4\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)
PTHH:
\(H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2H_2O\) (3)
\(Al_2\left(SO_4\right)_3+3Ba\left(OH\right)_2\rightarrow3BaSO_4\downarrow+2Al\left(OH\right)_3\downarrow\) (4)
\(FeSO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+Fe\left(OH\right)_2\downarrow\) (5)
Theo PT (3), (4), (5): \(n_{Ba\left(OH\right)_2\left(pư\right)}=m_{H_2SO_4\left(dư\right)}+n_{FeSO_4}+3n_{Al_2\left(SO_4\right)_3}=0,4\left(mol\right)< 0,45\left(mol\right)=n_{Ba\left(OH\right)_2\left(bđ\right)}\)
=> Ba(OH)2 dư
Theo PT (3), (4), (5): \(n_{BaSO_4}=n_{Ba\left(OH\right)_2\left(pư\right)}=0,4\left(mol\right)\)
Theo (4): \(n_{Al\left(OH\right)_3}=2n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\)
\(n_{Ba\left(OH\right)_2\left(dư\right)}=0,45-0,4=0,05\left(mol\right)\)
PTHH:
\(2Al\left(OH\right)_3+Ba\left(OH\right)_2\rightarrow Ba\left(AlO_2\right)_2+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{2}=\dfrac{0,05}{1}\) => Pư vừa đủ
\(4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^o}2Fe_2O_3+4H_2O\)
0,15------------------->0,075
=> m = 0,075.160 + 0,4.233 = 105,2 (g)
