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![](https://rs.olm.vn/images/avt/0.png?1311)
a,4FeS2 + 11O2 --->2 Fe2O3+ 4SO2
2SO2 +O2 --->2SO3
SO3 + H2O ---> H2SO4
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
b, \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
c, \(C_6H_{12}O_6+2AgNO_3+3NH_3\underrightarrow{t^o}C_5H_{11}O_5COONH_4+2Ag+2NH_4NO_3\)
d, \(\left(RCOO\right)_3C_3H_5+3NaOH\underrightarrow{t^o}3RCOONa+C_3H_5\left(OH\right)_3\)
e, \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
f, \(H_2SO_4+2CH_3COONa\rightarrow2CH_3COOH+Na_2SO_4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S+O_2\rightarrow\left(t^o\right)SO_2\)
\(2SO_2+O_2\rightarrow\left(t^o,V_2O_5\right)2SO_3\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(H_2SO_4+Mg\rightarrow MgSO_4+H_2\uparrow\)
\(MgSO_4+BaCl_2\rightarrow MgCl_2+BaSO_4\downarrow\)
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
\(Mg\left(OH\right)_2\rightarrow\left(t^o\right)MgO+H_2O\)
\(S+O_2\xrightarrow[]{t^o}SO_2\)
\(2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(H_2SO_4+Mg\left(OH\right)_2\rightarrow MgSO_4+H_2O\)
\(MgSO_4+BaCl_2\rightarrow BaSO_4\downarrow+MgCl_2\)
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
\(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ag + AgNO3 (không phản ứng)
Cu + 2AgNO3 -> 2Ag +Cu(NO3)2
0.04 0.08
n(Ag)=8.8/108=0.08 mol
m(Cu)=0.04*64=2.56(g)
%m(Cu)=(2.56*100)/5=51.2%
![](https://rs.olm.vn/images/avt/0.png?1311)
CuO+CO->Cu+CO2
x x x x
Fe2O3+3CO->2Fe+3CO2
y 3y 2y 3y
nCO=0.04(mol)
Ta có:x+3y=0.04
64x+112y=1.76
x=y=0.01(mol)
->a=0.01*80+0.01*160=2..4(g)
nCO2=0.01+0.01*3=0.04(mol)
V=0.896(l)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đổi 300ml = 0,3 lít
Ta có: \(n_{H_2SO_4}=5.0,3=1,5\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=\dfrac{\dfrac{8\%.400}{100\%}}{171}=\dfrac{32}{171}\left(mol\right)\)
a. PTHH: \(Ba\left(OH\right)_2+H_2SO_4--->BaSO_4\downarrow+2H_2O\)
Ta thấy: \(\dfrac{1,5}{1}>\dfrac{\dfrac{32}{171}}{1}\)
Vậy Ba(OH)2 dư, H2SO4 hết.
Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=1,5\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=1,5.233=349,5\left(g\right)\)
b. Không có muối tạo thành
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)O_2+S\xrightarrow[]{t^0}SO_2\\ 2SO_2+O_2\xrightarrow[]{t^0}2SO_3\\ SO_3+H_2O\rightarrow H_2SO_4\\ H_2SO_4+Na_2SO_3\rightarrow Na_2SO_4+SO_2+H_2O\\ SO_2+Na_2O\rightarrow Na_2SO_3\\ Na_2SO_3+Ba\left(NO_3\right)_2\rightarrow2NaNO_3+BaSO_3\\ b)4Na+O_2\rightarrow2Na_2O\\ Na_2O+H_2O\rightarrow2NaOH\\ NaOH+HCl\rightarrow NaCl+H_2O\\ NaOH+CO_{2,dư}\rightarrow NaHCO_3\\ 2NaHCO_3+H_2SO_4\rightarrow Na_2SO_4+2CO_2+2H_2O\\ c)Mg+2HCl\rightarrow MgCl_2+H_2\\ MgCl_2+Ca\left(OH\right)_2\rightarrow Mg\left(OH\right)_2+CaCl_2\\ Mg\left(OH\right)_2\xrightarrow[]{t^0}MgO+H_2O\\ MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ MgSO_4+Ba\left(NO_3\right)_2\rightarrow Mg\left(NO_3\right)_2+BaSO_4\\ Mg\left(NO_3\right)_2+Na_2CO_3\rightarrow MgCO_3+2NaNO_3\\ MgCO_3\xrightarrow[]{t^0}MgO+CO_2\\ CO_2+CaO\rightarrow CaCO_3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\((1)2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\\ (2)SO_3+H_2O\to H_2SO_4\\ (3)H_2SO_4+MgO\to MgSO_4 +H_2O\\ (4)MgSO_4+2NaOH\to Na_2SO_4+Mg(OH)_2\downarrow\)
1. S+O2->SO2(t0)
2. 2SO2+O2->2SO3(t0)
3. SO3+H2O->H2SO4
4. H2SO4+MgO->MgSO4+H2O