$V_A=0,1l$

$V_{H_2SO_4}=0,05l$

$\Rightarrow V_B=0,15l$

$[H^+]=10^{-1}(M)$

$\Rightarrow n_{H^+}=0,05.10^{-1}=0,005 mol$

Sau phản ứng, $[H^+]=10^{-3}$

$\Rightarrow n_{H^+\text{dư}}=10^{-3}.0,15=1,5.10^{-4}$

$\Rightarrow n_{NaOH}=n_{OH^-}=0,005-1,5.10^{-4}=4,85.10^{-3}$

Bảo toàn Na: $n_{Na}=n_{NaOH}$

$\to m=23.4,85.10^{-3}=0,11155g$

Bảo toàn e: $n_{H_2}=0,5n_{Na}$

$\Rightarrow V=4,85.10^{-3}.0,5.22,4=0,05432l$

nNaOH=$\frac{m}{40}$=0,025m (mol)

PT:              2Na+2H2O→2NaOH+H2

                  0,025m             0,025m  0,0125m        (mol)

=>nOH-=0,025m (mol)

nH+=2.nH2SO4=2.0,1.0,05=0,01 (mol)

Vdd sau pư=100+50=150ml=0,15 lít

pH=3<7=>axit dư=>H+ dư

=>[H+]dư=$\frac{0,01-0,025m}{0,15}$=10^-3

  =>m=0,394 (g)

nH2=0,0125m=4,925.10^-3 (mol)

=>V=VH2=4,925.10^-3.22,4=0,11032 (lít)