$\text{Đáp án+ Giải thích:}$

Bài 1:

$Pt: 3Zn + 2H_3PO_4$ $\rightarrow$ $Zn_3(PO_4)_2 + 3H_2 (1)$

$n_{H_2}$= $\frac{3,36}{22,4}$= $0,15(mol)$

$\text{Từ pt(1):}$ $n_{ZnCl_2}$= $0,05(mol)$

$\rightarrow$ $m_{Zn_3(PO_4)_2}$= $0,05. 385= 19,25(g)$

Bài 2:

$n_{Zn}$= $\frac{6,5}{65}$= $0,1(mol)$

$Pt: 3Zn + 2H_3PO_4$ $\rightarrow$ $Zn_3(PO_4)_2 + 3H_2 (1)$

$\text{Từ pt(1):}$ $n_{H_2}$= $0,1(mol)$

$\rightarrow$ $V_{H_2}$= $0,1.22,4= 2,24(l)$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
3Zn + 2{H_3}P{O_4} \to Z{n_3}{(P{O_4})_2} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Z{n_3}{{(P{O_4})}_2}}} = \dfrac{{{n_{{H_2}}}}}{3} = 0,05mol\\
{m_{Z{n_3}{{(P{O_4})}_2}}} = 0,05 \times 385 = 19,25g\\
2)\\
3Zn + 2{H_3}P{O_4} \to Z{n_3}{(P{O_4})_2} + 3{H_2}\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)