Bài1:
m ( Al)= (15,79.342):100=54(g)
m ( S)= (28,07.342):100=96(g)
m ( O)= 342-54-96=192(g)
=> n ( Al)= 54:27=2 (mol)
n (S)= 96:32=3 (mol)
n (O)= 192:16=12(mol)=4.3(mol)
--> CTHH của B là Al2(SO4)3
Bài 2:
Ta có:
d ( X: H2)= M (X) : M (H2)
(=) 8,5 = M (X) : 2
=) M (X)= 8,5.2=17
m (N)= (82,35.17):100=14(g)
m (H)= (17,65.17):100=3(g)
n (N)=14:14=1 (mol)
n (H)= 3:1= 3( mol)
--> NH3
Bài 3:
Ta có:
d ( X: KK)= M (X) : M (KK)
(=) 2,207 = M (X) : 29
=) M (X)= 29. 2,207=64
m (S)= (50.64):100=32(g)
m (O)= (50.64):100=32(g)
n (S)=32:32=1 (mol)
n (O)= 32: 16= 2 (mol)
--> SO2