Em tham khảo nha :

\(\begin{array}{l}
20)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
CuO + {H_2} \to Cu + {H_2}O\\
b)\\
{n_{Zn}} = \dfrac{{13}}{{65}} = 0,2mol\\
{n_{HCl}} = \dfrac{{10,95}}{{36,5}} = 0,3mol\\
\dfrac{{0,2}}{1} > \dfrac{{0,3}}{2} \Rightarrow Zn\text{ dư}\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,15mol\\
{n_{CuO}} = {n_{{H_2}}} = 0,15mol\\
{m_{CuO}} = 0,15 \times 80 = 12g\\
c)\\
{n_{Cu}} = {n_{{H_2}}} = 0,15mol\\
{m_{Cu}} = 0,15 \times 64 = 9,6g\\
21)\\
a)\\
{n_{{H_2}}} = \dfrac{{12 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 2mol\\
{n_{{O_2}}} = \dfrac{{9 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 1,5mol\\
{n_{hh}} = 2 + 1,5 = 3,5mol\\
{V_{hh}} = 3,5 \times 22,4 = 78,4l\\
b)\\
2{H_2} + {O_2} \to 2{H_2}O\\
\dfrac{2}{2} < \dfrac{{1,5}}{1} \Rightarrow {O_2}\text{ dư}\\
{n_{{H_2}O}} = {n_{{H_2}}} = 2mol\\
{m_{{H_2}O}} = 2 \times 18 = 36g
\end{array}\)

Bài 20:

$n_{Zn}=\dfrac{13}{65}=0,2(mol);n_{HCl}=\dfrac{10,95}{36,5}=0,3(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ CuO+H_2\xrightarrow{t^o}Cu+H_2O\\ \text{Lập tỉ lệ: }\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\Rightarrow Zn\text{ dư}\\ b,n_{CuO}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15(mol)\\ \Rightarrow m_{CuO}=0,15.80=12(g)\\ c,n_{Cu}=n_{CuO}=0,15(mol)\\ \Rightarrow m_{Cu}=0,15.64=9,6(g)$

Bài 21:

$a,n_{H_2}=\dfrac{12.10^{23}}{6.10^{23}}=2(mol)\\ n_{O_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5(mol)\\ \Rightarrow V_{hh}=V_{H_2}+V_{O_2}=2.22,4+1,5.22,4=78,4(l)\\ b,PTHH:2H_2+O_2\xrightarrow{\text{tia lửa điện}}2H_2O\\ \text{Lập tỉ lệ: }\dfrac{n_{H_2}}{2}<\dfrac{n_{O_2}}{1}\Rightarrow O_2\text{ dư}\\ \Rightarrow n_{H_2O}=n_{H_2}=2(mol)\\ \Rightarrow m_{H_2O}=2.18=36(g)$