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Bài 2.

a) \(\left(1\right)2HCl\rightarrow H_2+Cl_2\)

    \(\left(2\right)Cl_2+H_2\rightarrow2HCl\)

    \(\left(3\right)2Fe+3Cl_2\underrightarrow{t_o}2FeCl_3\)

    \(\left(4\right)2FeCl_3+Fe\rightarrow3FeCl_2\)

    \(\left(5\right)2NaOH+Cl_2\rightarrow H_2O+NaCl+NaClO\)

b) \(\left(1\right)2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(\left(2\right)Cl_2+H_2\rightarrow2HCl\)

\(\left(3\right)2HCl+FeO\rightarrow FeCl_2+H_2O\)

\(\left(4\right)3FeCl_2+4HNO_3\rightarrow2FeCl_3+Fe\left(NO_3\right)_3+NO+2H_2O\)

\(\left(5\right)2FeCl_3+Fe\rightarrow3FeCl_2\)

\(\left(6\right)FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\downarrow\)

Câu 1:

a)

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

\(Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\)

\(2KI+Cl_2\rightarrow2KCl+I_2\)

\(H_2+Cl_2\underrightarrow{as}2HCl\)

\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)

\(Cl_2+H_2O⇌HCl+HClO\)

b) 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)

\(HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

Câu 2:

a: (1): \(2HCl\rightarrow H_2+Cl_2\)(đk: điện phân dung dịch)

(2): \(Cl_2+H_2\rightarrow2HCl\)(đk: as)

(3): \(3Cl_2+2Fe\rightarrow2FeCl_3\)(điều kiện: t độ)

(4): \(2FeCl_3+Fe\rightarrow3FeCl_2\)(đk: đun sôi dung môi)

(5): \(Cl_2+2NaOH\rightarrow NaClO+NaCl+H_2O\)

b:

(1): \(2KMnO_4+16HCl\rightarrow5Cl_2+8H_2O+2KCl+2MnCl_2\)

(2): \(Cl_2+H_2\rightarrow2HCl\)

(3): \(2HCl+Fe\rightarrow FeCl_2+H_2\)

(4): \(2FeCl_2+Cl_2\rightarrow2FeCl_3\)

(5): \(2FeCl_3+Fe\rightarrow3FeCl_2\)

\(\left(6\right):FeCl_3+3AgNO_3\rightarrow3AgCl+Fe\left(NO_3\right)_3\)

BT e: \(3n_{Al}+2n_{Mg}=4n_{O_2}+2n_{Cl_2}\)

\(\Rightarrow3\cdot0,1+2\cdot0,05=4\cdot0,05+2\cdot x\)

\(\Rightarrow x=0,1mol\)

\(m=m_{Al}+m_{Mg}+m_{O_2}+m_{Cl_2}\)

    \(=0,1\cdot27+0,05\cdot24+0,05\cdot2\cdot16+0,1\cdot35,5\cdot2\)

    \(=12,6g\)

Theo bảo toàn electron ta có: \(3\cdot n_{Al}+2\cdot n_{Mg}=2\cdot n_{Cl_2}+4\cdot N_{O_2}\)

\(\Rightarrow3\cdot0,1+2\cdot0,05=4\cdot0,05+2x\Rightarrow x=0,2\)

\(\Rightarrow m_Z=m_X+m_Y=0,1\cdot27+0,05\cdot24+0,05\cdot32+0,2\cdot71=19,7g\)

\(C+O_2\underrightarrow{t_o}CO_2\)

\(N+O_2\underrightarrow{t_O}NO_2\)

\(4Na+O_2\underrightarrow{t_o}2Na_2O\)

\(2Ca+O_2\underrightarrow{t_O}2CaO\)

\(4Al+3O_2\underrightarrow{t_o}2Al_2O_3\)

\(C_2H_4+3O_2\underrightarrow{t_o}2CO_2+2H_2O\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t_o}2CO_2+H_2O\)

1: \(C+O_2\rightarrow CO_2\)

2: \(4N+O_2\rightarrow2N_2O\)

3: \(4Na+O_2\rightarrow2Na_2O\)

4 \(2Ca+O_2\rightarrow2CaO\)

5 \(4Al+3O_2\rightarrow2Al_2O_3\)

6: \(C_2H_4+O_2\rightarrow CH_3CHO\)

7: \(2C_2H_2+5O_2\rightarrow4CO_2+2H_2O\)

1.\(CO_2\)

2.\(NO_2\)

3.4Na+\(O_2\)->2\(Na_2O\)

4.CaO

5.\(4Al+3O_2->2Al_2O_3\)

6.\(C_2H_4+3O_2->2H_2O+2CO_2\)

7.\(2C_2H_2+5O_2->4CO_2+2H_2O\)

\(2HgO\rightarrow\left(450^oC-500^oC\right)2Hg+O_2\uparrow\)

C₂H₄+3O₂-to>2CO₂+2H₂O

              0,3-------0,2-------0,2

n _C2H4=\(\dfrac{4,48}{22,4}\)=0,2 mol

n _O2=\(\dfrac{6,72}{22,4}\)=0,3 mol

=>C2H4 dư

=>V_CO2=V_H2O=0,2.22,4=4,48l

=>V_{C2H4 dư}=0,1.22,4=2,24l

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ Vì:\dfrac{0,2}{1}>\dfrac{0,3}{3}\Rightarrow C_2H_4dư\\ \Rightarrow n_{CO_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ n_{C_2H_4\left(dư\right)}=0,2-\dfrac{0,3}{3}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ V_{C_2H_4\left(dư\right)}=0,1.22,4=2,24\left(l\right)\)

CuO+H₂-to>Cu+H₂O

0,1---------------0,1 mol

n _CuO=\(\dfrac{8}{80}\)=0,1 mol

H=80%

=>m _Cu=0,1.64.\(\dfrac{80}{100}\)=5,12g

CuO+H2\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)Cu+H2O

Theo PT: \(n_{H_{2\left(lt\right)}}\)=\(n_{CuO}\)=\(\dfrac{8}{80}\)=0,1(mol)

Vì hiệu suất đạt 80%

⇒\(V_{H_2}\)(thực)=\(\dfrac{0,1.22,4}{80\%}\)=2,8(l)

2KMnO₄-to>K₂MnO₄+MnO₂+O₂

0,2---------------------------------------0,1 mol

n _O2=\(\dfrac{2,24}{22,4}\)=0,1 mol

=>m _{KMnO4 tt} =0,2.158=31,6g

=>H =\(\dfrac{31,6}{39,5}.100\)=80%

Ta có: \(n_{KMnO_4}=\dfrac{39,5}{158}=0,25\left(mol\right)\)

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{O_2\left(LT\right)}=\dfrac{1}{2}n_{KMnO_4}=0,125\left(mol\right)\)

Mà: \(n_{O_2\left(TT\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow H\%=\dfrac{0,1}{0,125}.100\%=80\%\)

Bạn tham khảo nhé!

Tham khảo

\(M=27+\left(14+16.3\right).3=213\left(\dfrac{g}{mol}\right)\)

\(m=n.M=0,15.213=31,95\left(g\right)\)

Đề bài hỏi gì nhỉ?

\(m_{Al\left(NO_3\right)_3}=0,15.213=31,95\left(g\right)\)

\(V_{Al\left(NO_3\right)_3}=0,15.22,4=3,36\left(l\right)\)

mk k bt đề bài hỏi j nên mk làm 2 cách nha

\(a,Đặt:n_{Mg}=g\left(mol\right);n_{Fe}=j\left(mol\right)\left(g,j>0\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24g=56j=9,2\\22,4g+22,4j=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}g=0,15\\j=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\approx39,13\%\\ \Rightarrow\%m_{Fe}\approx60,87\%\\ b,n_{HCl}=2n_{H_2}=\dfrac{2.5,6}{22,4}=0,5\left(mol\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\ c,m_{muối}=m_{FeCl_2}+m_{MgCl_2}=95g+127j=95.0,15+127.0,1=26,95\left(g\right)\)

Fe+2HCl->FeCl₂+H₂

x-----------------------x mol

Mg+2HCl->MgCl₂+H₂

y-------------------------y mol

ta có\(\left\{{}\begin{matrix}56x+24y=9,2\\x+y=0,25\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)

=>%m_Fe=\(\dfrac{0,1.56}{9,2}.100\)=60,87%

=>%m _Mg=39,13%

Ta có : n _HCl=0,1.2+0,15.2=0,5 mol

=>CM_HCl=\(\dfrac{0,5}{0,2}\)=2,5M

=>m _muối =0,1.127+0,15.95=26,95g