`n_{CH_4}=\frac{3,136}{22,4}=0,14(mol)`
`n_{C_2H_2}=\frac{1,12}{22,4}=0,05(mol)`
`a)` `TH_1`:
Phương trình:
`Al_4C_3+12H_2O\to 4Al(OH)_3\downarrow+3CH_4`
`=> n_{Al_4C_3}=\frac{n_{CH_4}}{3}=\frac{0,14}{3}=\frac{7}{150}(mol)`
`=> m_1=\frac{7}{150}.144=6,72g`
`TH_2:`
`CaC_2+2H_2O\to C_2H_2+Ca(OH)_2`
`=> n_{CaC_2}=n_{C_2H_2}=0,05(mol)`
`=> m_2=0,05.64=3,2g`
`n_{Ca(OH)_2}=0,05(mol)`
`n_{Al(OH)_3}=\frac{4}{3}n_{CH_4}=\frac{14}{75}(mol)`
`b)` `Ca(OH)_2+2Al(OH)_3\to Ca(AlO_2)_2+4H_2O`
Do `\frac{0,05}{1}<\frac{\frac{14}{75}}{2}`
`n_{Ca(AlO_2)_2}=n_{Ca(OH)_2}=0,05(mol)`
`=> m_{Ca(AlO_2)_2}=0,05.158=7,9g`
`=> m_{\text{muối}}=7,9g`