Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = 0,5mol\\
 \to {n_{{H_2}}} = {n_{Zn}} = 0,5mol\\
 \to {V_{{H_2}}} = 11,2l\\
{n_{HCl}} = 2{n_{Zn}} = 1mol\\
 \to {m_{HCl}} = 36,5g\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = 0,52mol\\
 \to {n_{CuO}} > {n_{{H_2}}}
\end{array}\)

Suy ra CuO dư 

\(\begin{array}{l}
 \to {n_{Cu}} = {n_{{H_2}}} = 0,5mol\\
 \to {m_{Cu}} = 32g
\end{array}\)

2,

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = 0,075mol\\
 \to {n_{{H_2}}} = {n_{Mg}} = 0,075mol\\
 \to {V_{{H_2}}} = 1,68l\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,075mol\\
 \to {m_{MgC{l_2}}} = 7,125g\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = 0,1mol\\
 \to {n_{CuO}} > {n_{{H_2}}}
\end{array}\)

Suy ra CuO dư 

\(\begin{array}{l}
 \to {n_{Cu}} = {n_{{H_2}}} = 0,075mol\\
 \to {m_{Cu}} = 4,8g
\end{array}\)

3,

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = 0,25mol\\
 \to {n_{{H_2}}} = {n_{Fe}} = 0,25mol\\
 \to {V_{{H_2}}} = 5,6l\\
 \to {n_{HCl}} = 2{n_{Fe}} = 0,5mol\\
 \to {m_{HCl}} = 18,25g\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = 0,2mol\\
 \to {n_{CuO}} < {n_{{H_2}}}
\end{array}\)

Suy ra khí \({H_2}\) dư 

\(\begin{array}{l}
 \to {n_{Cu}} = {n_{CuO}} = 0,2mol\\
 \to {m_{Cu}} = 12,8g
\end{array}\)

Đáp án:

Bài 1:

$Zn + 2HCl→ZnCl_{2} + H_{2}↑$ (1)

$CuO + H_{2}\xrightarrow{t^o}Cu + H_{2}O$ (2)

a) Ta có: $n_{Zn}=\dfrac{m}{M}=\dfrac{32,5}{65}= 0,5 (mol)$

Theo PT (1): $n_{H_{2}}=n_{Zn}= 0,5 (mol)$

⇒ $V_{H_{2}}=n × 22,4=0,5 × 22,4= 11,2 (l)$

b) Theo PT (1): $n_{HCl}=2n_{Zn}=2 × 0,5= 1 (mol)$

⇒ $m_{HCl}=n × M=1 × 36,5= 36,5 (g)$

c) $n_{CuO}=\dfrac{m}{M}=\dfrac{41,6}{80}= 0,52 (mol)$

Lập tỉ lệ: $\dfrac{0,52}{1}>\dfrac{0,5}{1}$

⇒ $CuO$ dư, tính theo mol $H_{2}$

Theo PT (2): $n_{Cu}=n_{H_{2}}= 0,5 (mol)$

⇒ $m_{Cu}=n × M=0,5 × 64= 32 (g)$

Bài 2:

$Mg + 2HCl→MgCl_{2} + H_{2}↑$ (1)

$CuO + H_{2}\xrightarrow{t^o}Cu + H_{2}O$ (2)

a) Ta có: $n_{Mg}=\dfrac{m}{M}=\dfrac{1,8}{24}= 0,075 (mol)$

Theo PT (1): $n_{H_{2}}=n_{Mg}= 0,075 (mol)$

⇒ $V_{H_{2}}=n × 22,4=0,075 × 22,4= 1,68 (l)$

b) Theo PT (1): $n_{MgCl_{2}}=n_{Mg}= 0,075 (mol)$

⇒ $m_{MgCl_{2}}=n × M=0,075 × 95= 7,125 (g)$

c) $n_{CuO}=\dfrac{m}{M}=\dfrac{8}{80}= 0,1 (mol)$

Lập tỉ lệ: $\dfrac{0,1}{1}>\dfrac{0,075}{1}$

⇒ $CuO$ dư, tính theo mol $H_{2}$

Theo PT (2): $n_{Cu}=n_{H_{2}}= 0,075 (mol)$

⇒ $m_{Cu}=n × M=0,075 × 64= 4,8 (g)$

Bài 3:

$Fe + 2HCl→FeCl_{2} + H_{2}↑$ (1)

$CuO + H_{2}\xrightarrow{t^o}Cu + H_{2}O$ (2)

a) Ta có: $n_{Fe}=\dfrac{m}{M}=\dfrac{14}{56}= 0,25 (mol)$

Theo PT (1): $n_{H_{2}}=n_{Fe}= 0,25 (mol)$

⇒ $V_{H_{2}}=n × 22,4=0,25 × 22,4= 5,6 (l)$

b) Theo PT (1): $n_{HCl}=2n_{Fe}=2 × 025= 0,5 (mol)$

⇒ $m_{HCl}=n × M=0,5 × 36,5= 18,25 (g)$

c) $n_{CuO}=\dfrac{m}{M}=\dfrac{16}{80}= 0,2 (mol)$

Lập tỉ lệ: $\dfrac{0,2}{1}<\dfrac{0,25}{1}$

⇒ $H_{2}$ dư, tính theo mol $CuO$

Theo PT (2): $n_{Cu}=n_{CuO}= 0,2 (mol)$

⇒ $m_{Cu}=n × M=0,2 × 64= 12,8 (g)$