a) CuSO₄+FeFeSO+Cu

b) ZnSO+ 2NaOHNaSO+ Zn(OH)₂

c) HCl+AgNOAgCl+HNO

d) BaSO+SO

e)NaCO+Ca(NO)₂→ 2NaNO+ CaCO₃

Ca(H2PO4)2: muối axit: canxi đihiđrophotphat

NaHSO4: muối axit: natri hiđrosunfat

CaCO3: muối trung hòa: canxi cacbonat

Fe(OH)2: sắt (II) hiđroxit

Mg(NO3)2: muối trung hòa: magie nitrat

FeS: muối trung hòa: sắt (II) sunfua

NaCl: muối trung hòa: natri clorua

BaCl2: muối trung hòa: bari clorua

CuSO4: muối trung hòa: đồng (II) sunfat

Cu(HSO4)2: muối axit: đồng (II) hiđrosunfat

Cu(H2PO4)2: muối axit: đồng (II) đihiđrophotphat 

Muối: KNO₃, FeCl₂, Na₂CO₃, NaHCO₃

Axit: HCl

Bazơ: Mg(OH)₂

1. \(6NaOH+Fe_2\left(SO_4\right)_3\rightarrow2Fe\left(OH\right)_3+3Na_2SO_4\)

2. \(Mg+2AgNO_3\rightarrow Mg\left(NO_3\right)_2+2Ag\)

3.\(4Na+O_2\rightarrow2Na_2O\)

4.\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

chúc bn học tốt

1. 6NaOH + Fe₂(SO₄)₃ ➝ 2Fe(OH)₃ + 3Na₂SO₄

2. Mg + 2AgNO₃ ➝ Mg(NO₃)₂ + 2Ag

3. 4Na + O₂ \(\underrightarrow{t^o}\) 2Na₂O

4. 2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

\(1) Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O \text{Theo PTHH }\\ n_{H_2O} = n_{H_2} = \dfrac{20,16}{22,4}=0,9(mol)\\ \text{Bảo toàn khối lượng : }\\ a = m_{hh} + m_{H_2} - m_{H_2O} = 65,4 + 0,9.2 - 0,9.18 = 51(gam)\)

2)

\(n_{Mg} = a ; n_{Al} = b ; n_{Fe} = c\\ \Rightarrow 24a + 27b + 56c = 18,6(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{14,56}{22,4}=0,65(2)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ n_{O_2} = \dfrac{7,84}{22,4} = 0,35\)

Ta có :

\(\dfrac{a + b + c}{0,5a + 0,75b + \dfrac{2}{3}c} = \dfrac{0,55}{0,35}(3)\\ (1)(2)(3) \Rightarrow a = 0,2 ; b = 0,2 ; c= 0,15\\ \%m_{Mg} = \dfrac{0,2.24}{18,6}.100\% = 25,81\%\\ \%m_{Al} = \dfrac{0,2.27}{18,6}.100\% = 29,03\%\\ \%m_{Fe} = 100\% - 25,81\% -29,03\% = 45,16\%\)

\(a,\%Na=\dfrac{23}{85}.100\%=27,06\%\\ \%N=\dfrac{14}{85}.100\%=16,47\%\\ \%O=100\%-27,06\%-16,47\%=56,47\%\\ b,\%Al=\dfrac{54}{234}.100\%=27,1\%\\ \%C=\dfrac{36}{234}.100\%=15,4\%\\ \%O=100\%-27,1\%-15,4\%=57,5\%\)

\(c,\%N=\dfrac{28}{79}.100\%=35,4\%\\ \%H=\dfrac{4}{79}.100\%=5,1\%\\ \%O=100\%-35,4\%-5,1\%=59,5\%\)

\(M_{NaNO_3}=23+14+16.3=85\left(\dfrac{g}{mol}\right)\\ \Rightarrow\%m_{Na}=\dfrac{23.100\%}{85}=27\%\\ \%m_N=\dfrac{14.100\%}{85}=16,47\%\\ \Rightarrow\%m_O=100\%-\left(16,47\%+27\%\right)=56,53\)

$\rm u)S + 2HNO_3 \rightarrow H_2SO_4 + 2NO \uparrow$

$\rm v)3Ba(NO_3)_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 \downarrow + 2Al(NO_3)_3$
$\rm w)3Ba(OH)_2 + P_2O_5 \rightarrow Ba_3(PO_4)_3 \downarrow + 3H_2O$
$\rm x)2Fe_2O_3 + 3C \xrightarrow{t^o} 4Fe + 3CO_2 \uparrow$

\(1\\ 2Al\left(OH\right)_3\rightarrow\left(t^o\right)Al_2O_3+3H_2O\\ 2\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow3BaSO_4+2AlCl_3\)