a) \(Na_2O+H_2O\rightarrow2NaOH\)
b) \(n_{Na_2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)
Theo PTHH: \(n_{NaOH}=2.n_{Na_2O}=0,3\left(mol\right)\)
=> mNaOH = 0,3.40 = 12 (g)
=> \(C\%_{dd.NaOH}=\dfrac{12}{300}.100\%=4\%\)
c)
PTHH: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
0,3----->0,15
\(\Rightarrow V_{CO_2}=0,15.22,4=3,36\left(l\right)\)