$\color{Red}{\fbox{★Hóa 9★}}\color{Magenta}{\fbox{game nối chuỗi phản ứng }}$

[phamvananh9]

$CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O$
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[TEX][/TEX]
$Na_2CO_3 + BaCl_2--> 2NaCl + BaCO_3.$
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[kobato_2509]

$BaCO_3 + CuSO_4 \rightarrow BaSO_4 + CuCO_3$
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[lisel]

CuCO3 + H2SO4 ===> CuSO4 + H20 +CO2

 

[kobato_2509]

$CuSO_4 + Ba(NO_3)_2 \rightarrow Cu(NO_3)_2 + BaSO_4$
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[lisel]

$CuSO_4 + Ba(NO_3)_2 \rightarrow Cu(NO_3)_2 + BaSO_4$
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Cu (NO3) 2 + H2SO4 = CuSO 4 * 5H2O + H (NO3):khi (66)::khi (66)::khi (66):

 

[kobato_2509]

BaSO4 + H2O = H2SO4 + BaO:khi (66)::khi (66)::khi (66):

Không có cái này đâu bạn ơi
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[buivanbao123]

$BaSO_4+Pbcl_2------>BaCL_2+PbSO_4$

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[kagaminex]

$2PbSO_4---t^o--->2PbO+O_2+2SO_2$
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[lisel]

h2s + 3/2o2 ===> so2 + h2o@};-@};-@};-@};-@};-@};-@};-@};-

 

[buivanbao123]

$h_2o+k------------>koh+h_2$

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[kobato_2509]

$H_2 + Cl_2 \rightarrow 2HCl$
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[kagaminex]

$HCl+NaOH--->NaCl+H_2O$
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[kobato_2509]

$ H_2O + CaO \rightarrow Ca(OH)_2$
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[kagaminex]

$Ca(OH)_2+CO_2--->CaCO_3+H_2O$
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[kobato_2509]

$CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + H_2O + CO_2$
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[lisel]

CO2 + Ba(OH)2 = BaCO3 + H2O@};-@};-@};-@};-@};-@};-@};-

 

[kobato_2509]

CaSO4 + 2HCl ===> H2SO4 + CaCl2@};-@};-@};-@};-@};-@};-@};-

Khong tồn tại
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[qiana]

CaSO4 + 2HCl ===> H2SO4 + CaCl2@};-@};-@};-@};-@};-@};-@};-

Không thoả mãn điều kiện của phản ứng giữa muối với axit, các sp đều tan không có chất nào bị cách li khỏi dd sau phản ứng \Rightarrow không có phản ứng này.

 

[kagaminex]

$CaSO_4+Ba(NO_3)_2-----> BaSO_4+Ca(NO_3)_2$
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[kobato_2509]

$ BaSO_4 + 4C ---> BaS + 4CO$
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