Đáp án:
Giải thích các bước giải:
`a.`
`Al_xCl_y`
`n_{Al}=\frac{5,4}{27}=0,2(mol)`
`n_{Cl}=\frac{21,3}{35,5}=0,6(mol)`
`x:y=0,2:0,6`
`x:y=1:3`
`=>AlCl_3`
`b.`
`%m_O=100%-40%=60%`
`S_xO_y`
`x:y=40/32:60/16`
`x:y=1,25:3,75`
`x:y=1:3`
`=>SO_3`
`c.`
`%m_H=100%-75%=25%`
`C_xH_y`
`x:y=75/12:25/1`
`x:y=6,25:25`
`x:y=1/4`
`=>x=1/4y`
`M_{C_xH_y}=16(g//mol)`
`=>12x+y=16`
`=>12. 1/4y+y=16`
`=>y=4`
`=>x=1/4y=1`
`=>CH_4`