Question

cho 13,7 gam Ba vao 200 ml dung dich Al2(SO4)3 0,2M , ket thuc thu duoc m gam ket tua .Gia tri cua m la

Answer 1

nBa = 0,1 (mol)

Ba + 2H2O ---> Ba(OH)2 + H2 (1)

0,1........................0,1

nAl2(SO4)3 = 0,04 (mol)

3Ba(OH)2 + Al2(SO4)3 ---> 2Al(OH)3 + 3BaSO4

0,1...............................................\(\dfrac{1}{15}\).............0,1

Kết tủa là Al(OH)3 và BaSO4

\(\Rightarrow m=\dfrac{1}{15}.78+0,1.233=28,5\left(g\right)\)

Answer 2

+) Ta có: \(V_{ddAl_2\left(SO_4\right)_3}=200\left(ml\right)=0,2\left(l\right)\\ =>n_{Al_2\left(SO_4\right)_3}=0,2.0,2=0,04\left(mol\right)\\ n_{Ba}=\dfrac{13,7}{137}=0,1\left(mol\right)\\ PTHH:3Ba+Al_2\left(SO_4\right)_3->3BaSO_4\downarrow+2Al\\ \dfrac{0,1}{3}< \dfrac{0,04}{1}\)

=> Al₂(SO₄)₃ dư, Ba hết tính theo n_Ba.

Ta có: \(n_{BaSO_4}=n_{Ba}=0,1\left(mol\right)\\ =>m_{BaSO_4}=0,1.233=23,3\left(g\right)\)