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![](https://rs.olm.vn/images/avt/0.png?1311)
\(1) 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ 2) Na_2O + H_2SO_4 \to Na_2SO_4 + H_2O\\ 3) ZnO + H_2SO_4 \to ZnSO_4 + H_2O\\ 4) NaOH + HCl \to NaCl + H_2O\\ 5) 2NaOH + CO_2 \to Na_2CO_3 + H_2O\\ 6) Mg + 2HCl \to MgCl_2 + H)2\\ 7) Cu + 2AgNO_3 \to Cu(NO_3)_2 + 2Ag\\ 8) 2K + S \xrightarrow{t^o} K_2S\)
1, $2Fe(OH)_3\xrightarrow{t^o} Fe_2O_3+3H_2O$
2, $H_2SO_4+2NaOH\to Na_2SO_4+2H_2O$
3, $H_2SO_4+Zn(OH)_2\to ZnSO_4+2H_2O$
4, $NaOH+HCl\to NaCl+H_2O$
5, $2NaOH+CO_2\to Na_2CO_3+H_2O$
6, $Mg+2HCl\to MgCl_2+H_2$
7, $Cu+2AgNO_3\to Cu(NO_3)_2+2Ag$
8, $2K+S\xrightarrow{t^o} K_2S$
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
2) Al + 4HNO3 → Al(NO3)3 + NO + 2H2O
3) Al + 6HNO3 → Al(NO3)3 + 3NO2 + 3H2O
4) 2Fe + 6H2SO4 → Fe2(SO4)3 + 3SO2 + 6H2O
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,CH_4+Cl_2\underrightarrow{\text{ánh sáng}}CH_3Cl\\ 2,C_6H_6+Br_2\xrightarrow[t^o]{Fe}C_6H_5Br\\ 3,C_6H_6+Cl_2\underrightarrow{\text{ánh sáng}}C_6H_5Cl\\ 4,2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+CO_2\uparrow+H_2O\\ 5,\left(RCOO\right)_3C_3H_5+3NaOH\rightarrow3RCOONa+C_3H_5\left(OH\right)_3\\ 6,CaCO_3+CO_2+H_2O\rightarrow Ca\left(HCO_3\right)_2\\ 7,2NaHCO_3\underrightarrow{t^o}Na_2CO_3+CO_2\uparrow+H_2O\\ 8,NaOH+SiO_2\underrightarrow{t^o}Na_2SiO_3+H_2O\)
\(9,NaHCO_3+HCl\rightarrow NaCl+CO_2\uparrow+H_2O\\ 10,Cl_2+H_2O⇌HCl+HClO\\ 11,CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\\ 12,C_6H_6+3H_2\xrightarrow[t^o]{Ni}C_6H_{12}\\ 13,\left(RCOO\right)_3C_3H_5+3H_2O\rightarrow3RCOOH+C_3H_5\left(OH\right)_3\\ 14,\left(-C_6H_{10}O_5-\right)_n+nH_2O\rightarrow nC_6H_{12}O_6\\ 15,CH_3COONa+NaOH\underrightarrow{t^o}CH_4\uparrow+Na_2CO_3\\16, Ca\left(HCO_3\right)\underrightarrow{t^o}CaCO_3+CO_2\uparrow+H_2O\)
\(17,MnO_2+4HCl\rightarrow MnCl_2+Cl_2\uparrow+2H_2O\\ 18,NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\\ 19,2C_4H_{10}+5O_2\xrightarrow[men,xt]{t^o}4CH_3COOH+2H_2O\\ 20,2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\\ 21,CH_3COOC_2H_5+H_2O\underrightarrow{t^o}CH_3COOH+C_2H_5OH\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(a.PTHH:CO_2+Ba\left(OH\right)_2--->BaCO_3\downarrow+H_2O\)
b. Theo PT: \(n_{Ba\left(OH\right)_2}=n_{BaCO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{\dfrac{200}{1000}}=1M\)
c. Ta có: \(m_{BaCO_3}=0,2.197=39,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 :
\(n_{K_2O}=\dfrac{2.35}{94}=0.025\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
\(0.025...................0.05\)
\(C_{M_{KOH}}=\dfrac{0.05}{0.4}=0.125\left(M\right)\)
1) P2O5+H2O−>H3PO4
10) Na2O+H2O−>NaOH
2) CO2+H2O−>H2CO3
11) CaO + H2O−>Ca(OH)2
3) SO2+H2O−>H3SO3
12) BaO + H2O−>Ba(OH)2
4) SO3+H2O−>H2SO4
13) K2O+H2O−>.KOH
5) N2O5+H2O−>HNO3
14) Fe2O3+HCL−>FeCl3+H20
6) ......H3PO4.... + NaOH -> Na3PO4+H2O
15) ......MgO..+ H2SO4−>MgSO4+H2O
7) .CO2+ Ca(OH)2−>CaCO3+H2O
16) .KOH.....+H3PO4−>K3PO4+H2O
8) SO2...+ KOH -> K2SO3+H2O
17) .CaO+.CO2 ->CaCO3
9) .HNO3.+ Ba(OH)2−>Ba(NO3)2+H2O
18) .H3PO4.+..NaOH -> Na3PO4
thaks b nha