\(Zn+H_2SO_4->ZnSO_4+H_2\)

phản ứng thế

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) - Pư thế

\(2H_2O-^{nhietphan}>2H_2=+O_2\)

phản ứng phân hủy

a, PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Ta có: \(n_{C_2H_2}=\dfrac{13}{26}=0,5\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{C_2H_2}=1\left(mol\right)\Rightarrow m_{CO_2}=1.44=44\left(g\right)\)

\(n_{H_2O}=n_{C_2H_2}=0,5\left(mol\right)\Rightarrow m_{H_2O}=0,5.18=9\left(g\right)\)

b, Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=1,25\left(mol\right)\Rightarrow V_{O_2}=1,25.22,4=28\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=140\left(l\right)\)

c, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=1\left(mol\right)\Rightarrow m_{CaCO_3}=1.100=100\left(g\right)\)

n C2H2 = 1326=0,52613​=0,5 (mol)

2C2H2          +       3O2 ��→to​        4CO2 +     2H2O

0,5                     --> 0,75         --> 1         --> 0,5 

a. mCO2 = n.M = 1. 44 = 44 (gam) 

mH2O = n.M = 0,5 . 18 = 9 (gam) 

b. V(O2) = n.22,4 = 0,75 . 22,4 = 16,8 (lít) 

Ta có Oxi chiếm 20% thể tích không khí 

=> V (Không khí) = ��2.100%20%=16,8.10020=8420%VO2​​.100%​=2016,8.100​=84(lít) 

c. CO2 + Ca(OH)2 $\to$CaCO3 + H2O 

      1                         --> 1 

m CaCO3 = n.M = 1.100 = 100 (gam)

a, PT: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b, Ta có: \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{MgCO_3}=0,1.84=8,4\left(g\right)\)

\(\Rightarrow m_{MgO}=10,4-m_{MgCO_3}=2\left(g\right)\)

Ta có: \(\%m_C=\dfrac{12.13}{12.13+1.21+14+16.3}.100\%\approx65,272\%\)

Phần trăm khối lượng nguyên tử cacbon có trong phân tử Salbutamol là 

C% 12.1312.13+1.21+14+16.3.100%=15,06%12.13+1.21+14+16.312.13​.100%=15,06%

\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\)

\(Cl_2+H_2\underrightarrow{t^o}2HCl\)

\(HCl+NaOH\rightarrow NaCl+H_2O\)

\(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2+H_2\)

MnO2 + 4HCl $\to$ MnCl2 + Cl2 + 2H2O 

Cl2 + H2 $\to$ 2HCl 

HCl + NaOH $\to$ NaCl + H2O 

2NaCl + 2H2O →���đ���đpddcmn​ 2NaOH + H2 + Cl2 

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Ta có: \(n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)\)

a, \(n_{H_2}=n_{Fe}=0,12\left(mol\right)\Rightarrow V_{H_2}=0,12.24,79=2,9748\left(l\right)\)

b, \(n_{FeSO_4}=n_{Fe}=0,12\left(mol\right)\Rightarrow m_{FeSO_4}=0,12.152=18,24\left(g\right)\)

c, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,08\left(mol\right)\Rightarrow m_{Fe}=0,08.56=4,48\left(g\right)\)

a, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Ta có: \(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,025\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,025.22,4}{2,8}.100\%=20\%\\\%V_{CH_4}=100-20=80\%\end{matrix}\right.\)

b, Ta có: \(n_{CH_4}=\dfrac{2,8.80\%}{22,4}=0,1\left(mol\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,15\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)\)

\(\Rightarrow a=m_{CaCO_3}=0,15.100=15\left(g\right)\)

a, PT: �2�4+��2→�2�4��2C2​H4​+Br2​→C2​H4​Br2​

Ta có: ���2=4160=0,025(���)nBr2​​=1604​=0,025(mol)

Theo PT: ��2�4=���2=0,025(���)nC2​H4​​=nBr2​​=0,025(mol)

⇒{%��2�4=0,025.22,42,8.100%=20%%���4=100−20=80%⇒⎩⎨⎧​%VC2​H4​​=2,80,025.22,4​.100%=20%%VCH4​​=100−20=80%​

b, Ta có: ���4=2,8.80%22,4=0,1(���)nCH4​​=22,42,8.80%​=0,1(mol)

PT: ��4+2�2��→��2+2�2�CH4​+2O2​to​CO2​+2H2​O

�2�4+3�2��→2��2+2�2�C2​H4​+3O2​to​2CO2​+2H2​O

Theo PT: ���2=���4+2��2�4=0,15(���)nCO2​​=nCH4​​+2nC2​H4​​=0,15(mol)

PT: ��2+��(��)2→����3↓+�2�CO2​+Ca(OH)2​→CaCO3↓​+H2​O

Theo PT: �����3=���2=0,15(���)nCaCO3​​=nCO2​​=0,15(mol)

⇒�=�����3=0,15.100=15(�)⇒a=mCaCO3​​=0,15.100=15(g)

(1) \(2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\)

(2) \(C_2H_2+H_2\underrightarrow{t^o,Pd}C_2H_4\)

(3) \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

 (4) \(CH_4+Cl_2\underrightarrow{as}CH_3Cl+HCl\)

 (5) \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

 (6) \(C_2H_2+2H_2\underrightarrow{t^o,Ni}C_2H_6\)

 (7) \(C_2H_4+H_2\underrightarrow{t^o,Ni}C_2H_6\)

(1) 2��4→���1500���2�2+3�22CH4​1500oClln​C2​H2​+3H2​

(2) �2�2+�2��,��→�2�4C2​H2​+H2​to,Pd​C2​H4​

(3) �2�4+��2→�2�4��2C2​H4​+Br2​→C2​H4​Br2​

 (4) ��4+��2��→��3��+���CH4​+Cl2​as​CH3​Cl+HCl

 (5) �2�2+2��2→�2�2��4C2​H2​+2Br2​→C2​H2​Br4​

 (6) �2�2+2�2��,��→�2�6C2​H2​+2H2​to,Ni​C2​H6​

 (7) �2�4+�2��,��→�2�6C2​H4​+H2​to,Ni​C2​H6​