Đáp án:

$V_{H_2}$ = $2,479$$(l)$

$C_M$ = $1$$(M)$

$V_{NaOH}$ = $1$$(l)$

Giải thích các bước giải:

$a,$

$nZn$ = $\dfrac{6,5}{65}$ = $0,1$$(mol)$

$\text{Zn + 2$CH_3COOH$ → (CH_3COO)_2Zn + $H_2$}$

$b,$

-Theo phương trình: $nH_2$ = $nZn$ = $0,1$$(mol)$

-Vậy $V_{H_2}$ = $0,1$ * $24,79$ = $2,479(lít)$

-Cụng theo phương trình: $nCH_3COOH$ = $2nZn$ = $0,2$$(mol)$

$⇒$ $C_M$ $CH_3COOH$ = $\dfrac{0,2}{0,2}$ = $1$$(M)$

$c,$

$CH_3COOH$ + $NaOH$ → $CH_3COONa$ + $H_2O$

-Theo phương trình: $nNaOH$ = $nCH_3COOH$ = $0,2$$(mol)$

$→$ $V_{NaOH}$= $\dfrac{n}{C_M}$ = $\dfrac{0,2}{0,2}$ = $1$$(lít)$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1\,mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 24,79 = 2,479l\\
b)\\
{n_{C{H_3}COOH}} = 2{n_{Zn}} = 0,2\,mol\\
{C_M}C{H_3}COOH = \dfrac{{0,2}}{{0,2}} = 1M\\
c)\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{NaOH}} = {n_{C{H_3}COOH}} = 0,2\,mol\\
{V_{{\rm{dd}}NaOH}} = \dfrac{{0,2}}{{0,2}} = 1l
\end{array}\)