a) 2Na + 2H2O --> 2NaOH + H2
b) mH2O = 100.1 = 100 (g)
\(n_{Na}=\dfrac{3,45}{23}=0,15\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,15----------->0,15-->0,075
\(C_{M\left(NaOH\right)}=\dfrac{0,15}{0,1}=1,5M\)
c) mdd sau pư = 3,45 + 100 - 0,075.2 = 103,3 (g)
=> \(C\%_{NaOH}=\dfrac{0,15.40}{103,3}.100\%=5,81\%\)