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1.
Na2O + SO2\(\rightarrow\)Na2SO3
2.
CaO + H2O \(\rightarrow\)Ca(OH)2
3.ko xảy ra
4.
Al2O3 + 6HNO3 \(\rightarrow\)2Al(NO3)3 + 3H2O
5.
2Fe(OH)3 + 3H2SO4 \(\rightarrow\)Fe2(SO4)3 + 6H2O
6.
ZnO + SO3 \(\rightarrow\)ZnSO4
7.
Fe + H2SO4 \(\rightarrow\)FeSO4 + H2
8.ko xảy ra
9.
CuO + 2HNO3 \(\rightarrow\)Cu(NO3)2 + H2O
10.
2NaOH + H2SO4 \(\rightarrow\)Na2SO4 + 2H2O
11.
Fe(OH)3 + 3HNO3 \(\rightarrow\)Fe(NO3)3 + 3H2O
12.
Cu(OH)2 + 2HCl \(\rightarrow\)CuCl2 + 2H2O
\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.Zn+2AgNO_3\rightarrow Zn\left(NO_3\right)_2+2Ag\\ c.2Na+S\xrightarrow[]{t^0}Na_2S\\ d.Ca+Cl_2\xrightarrow[]{t^0}CaCl_2\\ e.MgO+2HNO_3\rightarrow Mg\left(NO_3\right)_2+H_2O\\ f.Fe+2HCl\rightarrow FeCl_2+H_2\\ g.CuO+2HCl\rightarrow CuCl_2+H_2O\\ h.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ i.Cl+2NaOH\rightarrow NaCl+NaClO+H_2O\\ k.MgO_2+4HCl_{đặc}\xrightarrow[nhẹ]{đun}MgCl_2+Cl_2+2H_2O\)
1. MgO + 2HNO\(_{3}\) -> Mg(NO\(_{3})_{2}\) +H\(_{2}O\)
2. Cu(OH)\(_{2}\) + 2HCl -> CuCl\(_{2}\) + 2H\(_{2}\)O
3. Na\(_{2} \)CO\(_{3}\) + 2HCl -> 2NaCl + H\(_{2}\)O + CO\(_{2} \) (vì H\(_{2}\)CO\(_{3} \) là 1 axit yếu nên sẽ bị phân hủy thành H\(_{2}\)O + CO\(_{2}\))
4. Fe\(_{3} \)O\(_{4}\) + 8HCl -> FeCl\(_{2}\) + 2FeCl\(_{3}\) + 4H\(_{2}\)O
5. Zn + H\(_{2}\)SO\(_{4}\) -> ZnSO\(_{4}\) + H\(_{2}\)
6. Fe\(_{3}\)O\(_{4}\) + 4H\(_{2}\)SO\(_{4}\) -> FeSO\(_{4}\) + Fe(SO\(_{4}\))\(_{3}\) + 4H\(_{2}\)O
1.Hãy tính số mol có trong:
\(a.27,2\left(g\right)ZnCl_2\\
n_{ZnCl_2}=\dfrac{27,2}{136}=0,2\left(mol\right)\\
b.V_{O_2\left(đktc\right)}=11,2\left(l\right)\\
n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\
c.150\left(ml\right)ddNaOH2M\\
n_{NaOH}=0,15.2=0,3\left(mol\right)\\
d.200\left(g\right)ddH_2SO_419,6\%\\
n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)
Cho 2,7gam Al phản ứng với dd có chứa 29,4gam H2SO4.
a. Lập PTHH
b. Chất nào dư sau phản ứng và dư bao nhiêu gam.
c. Tính khối lượng muối thu được.
d. Tính thể tích khí sinh ra( đktc).
----
\(a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\b. Vì:\dfrac{0,1}{2}< \dfrac{0,3}{3}\Rightarrow H_2SO_4dư\\ m_{H_2SO_4\left(dư\right)}=98.\left(0,3-0,1.\dfrac{3}{2}\right)=14,7\left(g\right)\\ c.n_{Al_2\left(SO_4\right)_3}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=342.0,05=17,1\left(g\right)\\ d.n_{H_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)
1, CO2+H2O--->H2CO3
2, SO3+H2O--->H2SO4
3, SO2+H2O--->H2SO3
4, N2O5+H2O---> 2HNO3
5, P2O5+3H2O--->2H3PO4
6, SO2+K2O--->K2SO3
7, CO2+BaO--->BaCO3
8, SO3+Na2O--->Na2SO4
9, P2O5+3CaO--->Ca3(PO4)2
10, N2O5+K2O--->2KNO3
11, CO2+2NaOH--->Na2CO3+H2O
12, SO2+Ca(OH)2--->CaSO3+H2O
13, SO3+Ba(OH)2--->BaSO4+H2O
14, N2O5+2KOH--->2KNO3+H2O
15, P2O5+3Ba(OH)2--->Ba3(PO4)2+3H2O
16, Na2O+H2O--->2NaOH
17, K2O+H2O--->2KOH
18, CaO+H2O--->Ca(OH)2
19, BaO+H2O--->Ba(OH)2
20, Al2O3+6HCl--->2AlCl3+3H2O
21, Fe2O3+6HNO3---> 2Fe(NO3)3+3H2O
22, ZnO+H2SO4--->ZnSO4+H2O
23, 3CaO+2H3PO4--->Ca3(PO4)2+3H2O
24, Fe+2HCl--->FeCl2+H2
25, Mg+H2SO4--->MgSO4+H2
26, 2Al+6HCl--->2AlCl3+3H2
27, Zn+H2SO4---->ZnSO4+H2
28, Cu+2H2SO4---> CuSO4+SO2+2H2O
29, Al(OH)3+3HCl--->AlCl3+3H2O
30, Zn(OH)2+H2SO4--->ZnSO4+2H2O
\(a,MgO+2HNO_3\rightarrow Mg\left(NO_3\right)_2+H_2O\\ b,CuO+2HCl\rightarrow CuCl_2+H_2\\ c,Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ d,Fe+2HCl\rightarrow FeCl_2+H_2\\ e,Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a) Axit sunfuric + kẽm oxit → Kẽm sunfat + Nước
H2SO4 + ZnO → ZnSO4 + H2O
b) Natri hiđroxit + lưu huỳnh trioxit → Natri sunfat + Nước
2NaOH + SO3 → Na2SO4 + H2O
c) Nước + lưu huỳnh đioxit → Axit sunfurơ
H2O + SO2 → H2SO3
d) Nước + canxi oxit → Canxi hiđroxit
H2O + CaO → Ca(OH)2
e) Canxi oxit + cacbon đioxit→ Canxi cacbonat
CaO + CO2 → CaCO3