$n_{H_2}=2,24/22,4=0,1mol$
$a.2Na+2H_2O\to 2NaOH+H_2↑(1)$
$Na_2O+H_2O\to 2NaOH$
Theo pt (1) :
$n_{Na}=2.n_{H_2}=2.0,1=0,2mol$
$⇒m_{Na}=0,2.23=4,6g$
$⇒m_{Na_2O}=10,8-4,6=6,2g$
$b.n_{Na_2O}=6,2/62=0,1mol$
$n_{NaOH}=0,2+0,2=0,5mol$
$⇒m_{NaOH}=0,4.40=16g$
$c.C_{M_{NaOH}}=\dfrac{0,4}{0,5}=0,8M$
$m_{dd\ spu}=10,8+500-0,1.2=510,6g$
$⇒C\%_{NaOH}=\dfrac{16}{510,6}=3,13\%$
