CHƯƠNG II: PHẢN ỨNG HÓA HỌC

a, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

b, \(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)

c, \(H_2O+Na_2O\rightarrow2NaOH\)

Bạn tham khảo nhé!

\(A) Fe + 2HCl \to FeCl_2 + H_2\\ B) Ba(OH)_2 + 2HBr \to BaCl_2 + 2H_2O\)

\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)

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

\(0.05......0.1\)

\(m_{HCl}=0.1\cdot36.5=3.65\left(g\right)\)

Chỉ tìm được khối lượng HCl thoi em nhé !

Pthh

Mg+2HCl--->MgCl2+H2

2Al+6HCl--->2AlCl3+3H2

nAl=10,8/27=0,4 mol

nO2=3,36/22,4=0,15 mol

4Al + 3O2 --> 2Al2O3

 0,2   0,15           0,1   mol

ta thấy nAl/4=0,1 >nO2/3=0,05 

=> mAl2O3=0,05*102=5,1 g

mAl=0,2*27=5,4g

mAl sau p/ứ=10,8-5,4=5,4 g

\(n_{CO_2}=n_{CO}=\dfrac{p}{100}\left(mol\right)\)

\(\text{Áp dụng định luật bảo toàn khối lượng: }\)

\(m_X+m_{CO}=m_Y+m_{CO_2}\)

\(\Leftrightarrow m_X-m_Y=m_{CO_2}-m_{CO}\)

\(\Leftrightarrow\) \(m-n=\dfrac{p}{100}\cdot44-\dfrac{p}{100}\cdot28=0.16p\)

\(\Leftrightarrow m=n+0.16p\)

\(a) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ H_2O + SO_2 \to H_2SO_3\\ Mg + H_2SO_3 \to MgSO_3 + H_2\\ b) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ 2H_2O \xrightarrow{đp} 2H_2 +O_2\\ FeO + H_2 \xrightarrow{t^o} Fe + H_2O\\ Fe + 2HCl \to FeCl_2 + H_2\)

a) Cl2 + H2 --->2HCl

2HCl + 2Na ---> 2NaCl + H2

2H2 + O2 --to-> 2H2O

H2O + SO2 --> H2SO3

H2SO3  + Mg ---> MgSO3 + H2

b)2 KCLO3 --to-> 2KCl +3O2

O2 + 2H2 -to-> 2H2O

2H2O + Ca --> Ca(OH)2 + H2

H2 + FeO --> Fe +H2O

Fe + 2HCl --> FeCl2 + H2

\(a) 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ b) n_{Cu}=\dfrac{12,7}{64} = \dfrac{127}{640}(mol)\\ \Rightarrow n_{O_2} = \dfrac{1}{2}n_{Cu} = \dfrac{127}{1280}(mol)\\ \Rightarrow m_{O_2} = \dfrac{127}{1280}.32 = 3,175(gam)\\ c) V_{không\ khí} = 5V_{O_2} = 5.\dfrac{127}{1280}.22,4 = 11,1125(lít) \)