$\text{Đáp án+Giải thích các bước giải:}$

$\text{Bài 1:}$

$\text{PbO + $H_{2}$$\xrightarrow{t^o}$Pb + $H_{2}O$}$

$\text{A.}$

$\text{Ta có: $n_{Pb}$=$\dfrac{41,4}{207}$= 0,2 (mol)}$

$\text{Theo PT ta có: $n_{H_{2}}$=$n_{Pb}$= 0,2 (mol)}$

$\text{⇒ $V_{H_{2}}$=0,2 × 22,4= 4,48 (lít)}$

$\text{B.}$

$\text{Theo PT ta có: $n_{PbO}$=$n_{Pb}$= 0,2 (mol)}$

$\text{⇒ $m_{PbO}$=0,2 × 223= 44,6 (g)}$

$\text{Bài 2:}$

$\text{A.}$

$\text{Zn + 2HCl→$ZnCl_{2}$ + $H_{2}$}$

$\text{Phản ứng thế}$

$\text{B.}$

$\text{Ta có: $n_{Zn}$=$\dfrac{6,5}{65}$= 0,1 (mol)}$

$\text{Theo PT ta có: $n_{H_{2}}$=$n_{Zn}$= 0,1 (mol)}$

$\text{⇒ $V_{H_{2}}$=0,1 × 22,4= 2,24 (lít)}$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
PbO + {H_2} \to Pb + {H_2}O\\
{n_{Pb}} = \dfrac{{41,4}}{{207}} = 0,2mol\\
{n_{{H_2}}} = {n_{Pb}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{PbO}} = {n_{Pb}} = 0,2mol\\
{m_{PbO}} = 0,2 \times 223 = 44,6g\\
2)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
\\
b)\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)