`Ca+2H_2O->Ca(OH)_2+H_2↑`
`CaO+H_2O->Ca(OH)_2`
`n_{Ca(OH)_2}=(18,5)/74=0,25(mol)`
Gọi `a,b` là `n_{Ca},n_{CaO}`
`=>m_{hh}=40a+56b=12,08(1)`
`∑n_{Ca(OH)_2}=a+b=0,25(2)`
`(1),(2)=>a=0,12;b=0,13`
`n_{H_2}=n_{Ca}=a=0,12(mol)`
`V_{H_2}=0,12.22,4=2,688(l)`
`\text{___________________________________}`
Đáp án:
`V=2,688(l)`