Đáp án:
a) $Na_{2}O$ + $H_{2}O$ → $2NaOH $
$C_{M_{NaOH}}$=$\frac{2.0,25}{0,5}$=1M
b) $2NaOH$ + $H_{2}SO_{4}$ → $Na_{2}SO_{4}$ + $2H_{2}O$
$n_{H_{2}SO_{4}}$=$\frac{1}{2}$.$n_{NaOH}$=0,25(mol)
⇒$m_{H_{2}SO_{4}}$=0,25.98=24,5(g)
⇒$m_dd{H_{2}SO_{4}}$=$\frac{24,5}{20}.100$=122,5(g)
⇒$V_{H_{2}SO_{4}}$=$\frac{122,5}{1,14}$=107,46(ml)