a) \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
0,1----------------->0,2
b) \(C_{M\left(NaOH\right)}=\dfrac{0,2}{0,1}=2M\)
c) PTHH: 2NaOH + H2SO4 ---> Na2SO4 + 2H2O
0,2------>0,1
=> \(V_{ddH_2SO_4}=\dfrac{0,1}{2}=0,05\left(l\right)=50\left(ml\right)\)