`#Ly`

$a)$

PTHH: $2Na$ $+$ $H_2SO_4$ `->` $Na_2SO_4$ $+$ $H_2↑$

`n_(Na` `=` `m/M` `=` `(2,3)/(23)=0,1` $(mol)$

Theo PT, `n_(H_2` `=` `(1)/(2)n_(Na` `=` `0,05` $(mol)$

`V_(H_2(đktc)` `=` `n.22,4=0,05.22,4=1,12` $(l)$

$b)$

Theo PT, `n_(H_2SO_4` `=` `n_(H_2` `=` `0,05` $(mol)$

`m_(H_2SO_4` `=` `n.M=0,05.98=4,9` $(g)$

`C%_(H_2SO_4``=(m_{H_2SO_4})/(m_{ddH_2SO_4}).100%``=(4,9)/(600).100%=0,82%`

$c)$

`V_(H_2SO_4` `=` `n/(C_M)` `=` `(0,05)/(2)=0,025` $(l)$

$d)$

Theo PT, `n_(Na_2SO_4` `=` `n_(H_2` `=` `0,05` $(mol)$

`m_(Na_2SO_4` `=` `n.M=0,05.142=7,1` $(g)$

$e)$

`m_(dd(spư)` `=` `2,3+600-0,05.2=602,2` $(g)$

`C%_(Na_2SO_4=``(m_{Na_2SO_4})/(m_{ddNa_2SO_4}).100%``=(7,1)/(602,2).100%=1,18%`

$f)$

Do thể tích dd sau pư không đổi

`=>` `V_(dd` `=` `V_(H_2SO_4` `=` `0,025` $(l)$

`C_MNa_2SO_4` `=` `(n)/V` `=` `(0,05)/(0,025)=2` $(M)$

\(\begin{array}{l}
a)\\
2Na+H_2SO_4\to Na_2SO_4+H_2\\
n_{Na}=\frac{2,3}{23}=0,1(mol)\\
n_{H_2}=\frac{1}{2}n_{Na}=0,05(mol)\\
V_{H_2}=0,05.22,4=1,12(l)\\
b)\\
n_{H_2SO_4}=n_{H_2}=0,05(mol)\\
C\%_{H_2SO_4}=\frac{0,05.98}{600}.100\%\approx 0,817\%\\
c)\\
V_{dd\,H_2SO_4}=\frac{0,05}{2}=0,025(l)=25(ml)\\
d)\\
n_{Na_2SO_4}=n_{H_2}=0,05(mol)\\
m_{Na_2SO_4}=0,05.142=7,1(g)\\
e)\\
m_{dd\,spu}=2,3+600-0,05.2=602,2(g)\\
C\%_{Na_2SO_4}=\frac{7,1}{602,2}.100\%\approx 1,18\%\\
f)\\
V_{dd\,Na_2SO_4}=V_{dd\,H_2SO_4}=0,025(l)\\
C_{M\,Na_2SO_4}=\frac{0,05}{0,025}=2M
\end{array}\)