cứuuuuuuuuuuuuuuu với vote 5 sao nha

`a)`

Bảo toàn KL: `m_X+m_{H_2O}=m_{dd\ bazo}+m_{H_2}`

`->m_{H_2}=10,8+189,4-200=0,2(g)`

`->n_{H_2}={0,2}/2=0,1(mol)`

`->V=0,1.22,4=2,24(l)`

`b)`

TH1: M hóa trị `I`

`m_{MOH}=200.8\%=16(g)`

`2M+2H_2O->2MOH+H_2`

`M_2O+H_2O->2MOH`

Theo PT: `n_M=2n_{H_2}=0,2(mol)`

`->m_M=0,2M(g)`

`->m_{M_2O}=10,8-0,2M(g)`

`n_M+2n_{M_2O}=n_{MOH}`

`->0,2+2n_{M_2O}={16}/{M+17}`

`->0,2+{21,6-0,4M}/{2M+16}={16}/{M+17}`

`->M=23(Na)`

TH2: M hóa trị `II`

`m_{M(OH)_2}=200.8\%=16(g)`

`M+2H_2O->M(OH)_2+H_2`

`MO+H_2O->M(OH)_2`

Theo PT: `n_M=n_{H_2}=0,1(mol)`

`->m_M=0,1M(g)`

`->m_{MO}=10,8-0,1M(g)`

`n_M+n_{MO}=n_{M(OH)_2}`

`->0,1+n_{MO}={16}/{M+34}`

`->0,1+{10,8-0,1M}/{M+16}={16}/{M+34}`

`->M=46` (loại)

Vậy `M` là `Na,` oxit là `Na_2O`