2Na + 2H2O ->2NaOH + H2 (1)
Na2O +H2O -> 2NaOH (2)
nH2=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PTHH 1 ta có:
nNaOH=nNa=2nH2=0,4(mol)
mNa=23.0,4=9,2(g)
mNa2O=15,4-9,2=6,2(g)
nNa2O=6,2/62=0,1(mol)
Theo PTHH 2 ta có:
nNaOH=2nNa2O=0,2(mol)
=>\(\sum\)nNaOH=0,2+0,4=0,6(mol)
C% dd NaOH=\(\dfrac{0,6.40}{15,4+185-0,2.2}.100\%=12\%\)