Đáp án + Giải thích các bước giải:

Bài 12:

$a)PTHH:H_2+Cl_2\xrightarrow{t^o}2HCl$

$b)n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$

$n_{Cl_2}=\dfrac{6,72}{22,4}=0,3(mol)$

Xét tỉ lệ: $n_{H_2}>n_{Cl_2}\Rightarrow H_2$ dư

Theo PT: $n_{H_2(dư)}=0,4-0,3=0,1(mol)$

$\Rightarrow m_{H_2(dư)}=0,1.2=0,2(g)$

$c)PTHH:CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Theo PT: $n_{Cu}=n_{H_2(dư)}=0,1(mol)$

$\Rightarrow m_{Cu}=0,1.64=6,4(g)$

Bài 13:

$a)PTHH:CaCO_3+2HCl\to CaCl_2+H_2O+CO_2\uparrow$

$b)n_{CaCO_3}=\dfrac{10}{100}=0,1(mol)$

$n_{HCl}=\dfrac{5,475}{36,5}=0,15(mol)$

Xét tỉ lệ: $n_{CaCO_3}>\dfrac{n_{HCl}}{2}\Rightarrow CaCO_3$ dư

Theo PT: $n_{CaCO_3(dư)}=0,1-\dfrac{1}{2}.0,15=0,025(mol)$

$\Rightarrow m_{CaCO_3(dư)}=0,025.100=2,5(g)$

$c)$ Theo PT: $n_{CO_2}=\dfrac{1}{2}n_{HCl}=0,075(mol)$

$\Rightarrow V_{CO_2(đkct)}=0,075.22,4=1,68(l)$

$d)$ Để p/ứ vừa đủ

$\Rightarrow n_{CaCO_3}=\dfrac{n_{HCl}}{2}$

$\Rightarrow \dfrac{n_{HCl}}{2}=0,1$

$\Rightarrow n_{HCl}=0,2$

$\Rightarrow n_{HCl(thêm)}=0,2-0,15=0,05(mol)$

$\Rightarrow m_{HCl(thêm)}=0,05.36,5=1,825(g)$

Bài 14:

$a)PTHH:$

$2H_2+O_2\xrightarrow{t^o}2H_2O(1)$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4(2)$

$Fe+2HCl\to FeCl_2+H_2\uparrow(3)$

$Fe_3O_4+8HCl\to 2FeCl_3+FeCl_2+4H_2O(4)$

$A:H_2O$

$B:O_2$

$D:Fe$ dư và $Fe_3O_4$

$H:FeCl_3$ và $FeCl_2$

$E:H_2$

$b)n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$n_{O_2}=\dfrac{3,36}{22,4}=0,15(mol)$

$n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$

$n_{HCl}=\dfrac{14,6}{36,5}=0,4(mol)$

Xét tỉ lệ PT(1): $0,5n_{H_2}<n_{O_2}\Rightarrow O_2$ dư

Theo PT: $n_{H_2O}=n_{H_2(1)}=0,2(mol)$

$\Rightarrow m_A=m_{H_2O}=0,2.18=3,6(g)$

Theo PT: $n_{O_2(dư)}=0,15-\dfrac{1}{2}.0,2=0,05(mol)$

Xét tỉ lệ PT(2): $\dfrac{n_{Fe}}{3}>\dfrac{n_{O_2(1)}}{2}\Rightarrow Fe$ dư

Theo PT: $n_{Fe(dư)}=0,1-\dfrac{3}{2}.0,05=0,025(mol);n_{Fe_3O_4}=0,025(mol)$

Trong D: $\begin{cases} m_{Fe(dư)}=0,025.56=1,4(g)\\ m_{Fe_3O_4}=0,025.232=5,8(g) \end{cases}$

Xét tỉ lệ PT(4): $n_{Fe_3O_4}<\dfrac{n_{HCl}}{8}\Rightarrow HCl$ dư

Theo PT: $\Sigma n_{FeCl_2}=n_{Fe(dư)}+n_{Fe_3O_4}=0,05(mol);n_{FeCl_3}=2n_{Fe_3O_4}=0,05(mol)$

Trong H: $\begin{cases} m_{FeCl_2}=0,05.127=6,35(g)\\ m_{FeCl_3}=0,05.162,5=8,125(g) \end{cases}$