nH2 = 0,05 mol
=> nNa = 0,1 mol
=> mNa = 2,3 gam
=> mBaO = 7,65 gam
nNaOH = 0,1 mol
=> mNaOH = 4 gam
nBaO = 7,65/153 = 0,05 mol
=> nBa(OH)2 = 0,05 mol
=> mBa(OH)2 = 8,55 gam
- Theo PTHH(1,2); nH2O = 0,15 mol
=> mH2O = 2,7 gam = 2,7.10-3 (kg) = 2,7.10-3 lít