Cho sơ đồ pư sau
A \(\underrightarrow{HCl}\) B \(\underrightarrow{NaOH}\) D \(\underrightarrow{t^o}\) CuO
A là: A. Cu B. CuO C. Cu(OH)2 D. CuSO4
Viết các phương trình hoá học theo các sơ đồ chuyển hoá sau:
a) CuO \(\underrightarrow{+?}\) CuSO4 \(\underrightarrow{+?}\) Cu(OH)2 b) Mg \(\underrightarrow{+?}\) MgCl2 \(\underrightarrow{+?}\) Mg(OH)2
c) NaOH \(\underrightarrow{+?}\) Na2SO4 \(\underrightarrow{+?}\) NaCl c) K2CO3 \(\underrightarrow{+?}\) CaCO3 \(\underrightarrow{+?}\) CaCl2
a
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b
\(Mg+Cl_2\underrightarrow{t^o}MgCl_2\)
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
c
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl\)
d
\(K_2CO_3+CaCl_2\rightarrow CaCO_3+2KCl\)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
2, Hoàn thành phương trình theo sơ đồ sau :
Cu \(\underrightarrow{O^{ }_2}\) CuO \(\underrightarrow{+HCl}\) CuCl2 \(\underrightarrow{+Fe}\) Cu
3, Hòa tan 16g Fe2O3 vào 200g dung dịch H2SO4 19,6 % . Tính C% của chất tan sau phản ứng
*Hoàn thành ptpư theo sơ đồ :
\(Cu+\dfrac{1}{2}O_2\underrightarrow{t^o}CuO\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(CuCl_2+Fe\rightarrow Cu+FeCl_2\)
1)
2Cu + O2 --> 2CuO
CuO + 2HCl --> CuCl2 + H2O
CuCl2 + Fe --> Cu + FeCl2
3)
nH2SO4 = (200x19.6)/(98x100) = 0.4 (mol)
nFe2O3 = 16/160 = 0.1(mol)
PTHH Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
theo đề 0.1 0.4
đã pư 0.1 0.3
sau pư 0 0.1 0.1
Lập tỉ số nFe2O3 < nH2SO4
=> nH2SO4 Dư
mdd(sau pư) = 16 + 200 = 216 (g)
mFe2(SO4)3 = 0.1x400 = 40(g)
C%Fe2(SO4)3 = 40/216*100% = 18.518%
Viết các PTHH hoàn thành chuỗi chuyển hóa sau:
\(Cu\left(OH\right)_2\underrightarrow{\left(1\right)}CuO\underrightarrow{\left(2\right)}Cu\underrightarrow{\left(3\right)}CuCl_2\underrightarrow{\left(4\right)}Cu\left(OH\right)_2\underrightarrow{\left(5\right)}CuSO_4\)
(1) Cu(OH)2 -t0-> CuO + H2O
(2) CuO + H2 -t0-> Cu + H2O
(3) Cu + Cl2 -t0-> CuCl2
(4) CuCl2 + 2NaOH - > Cu(OH)2 + 2NaCl
(5) Cu(OH)2 + H2SO4 - > CuSO4 + 2H2O
Câu 1 ) A / Mg(NO3) \(\underrightarrow{\left(1\right)}\) Mg(OH)2 \(\underrightarrow{\left(2\right)}\) MgCl2\(\underrightarrow{\left(3\right)}\) KCl \(\underrightarrow{\left(4\right)}\) KNO3
B/ \(Na\underrightarrow{\left(1\right)}Na_2O\underrightarrow{\left(2\right)}NaOH\underrightarrow{\left(3\right)}Na_2SO_4\underrightarrow{\left(4\right)}NaCl\underrightarrow{\left(5\right)}NaNO_3\underrightarrow{\left(6\right)}NaCl\underrightarrow{\left(7\right)}NaOH\)
C/\(Mg\underrightarrow{\left(1\right)}MgO\underrightarrow{\left(2\right)}MgCl_2\underrightarrow{\left(3\right)}Mg\left(NO_3\right)_2\underrightarrow{\left(4\right)}Mg\left(OH\right)_2\underrightarrow{\left(5\right)}MgSO_4\underrightarrow{\left(6\right)}MgCO_3\)
D/\(CuSO_4\underrightarrow{\left(1\right)}Cu\left(OH\right)_2\underrightarrow{\left(2\right)}CuO\underrightarrow{\left(3\right)}CuCl_2\underrightarrow{\left(4\right)}Cu\left(OH\right)_2\underrightarrow{\left(5\right)}CuSO_4\)
E/\(CuCl_2\underrightarrow{\left(1\right)}Cu\left(OH\right)_2\underrightarrow{\left(2\right)}CuSO_4\underrightarrow{\left(3\right)}Cu\underrightarrow{\left(4\right)}CuO\)
F/ \(Cu\left(OH\right)_2\underrightarrow{\left(1\right)}CuO\underrightarrow{\left(2\right)}CuCl_2\underrightarrow{\left(3\right)}Cu\left(NO_3\right)_2\underrightarrow{\left(4\right)}NaNO_3\)
G/\(Fe_2O_3\underrightarrow{\left(1\right)}FeCl_3\underrightarrow{\left(2\right)}Fe\left(OH\right)_3\underrightarrow{\left(3\right)}Fe_2O_3\underrightarrow{\left(4\right)}Fe_2\left(SO_4\right)_3\)
H/ \(ZnCl_2\underrightarrow{\left(1\right)}Zn\left(OH\right)_2\underrightarrow{\left(2\right)}ZnCl_2\underrightarrow{\left(3\right)}NaCl\underrightarrow{\left(4\right)}NaNO_3\)
M/\(CuO\underrightarrow{\left(1\right)}CuCl_2\underrightarrow{\left(2\right)}Cu\left(OH\right)_2\underrightarrow{\left(3\right)}CuO\underrightarrow{\left(4\right)}CuSO_4\)
N/\(Fe\left(OH\right)_2\underrightarrow{\left(1\right)}FeO\underrightarrow{\left(2\right)}FeCl_2\underrightarrow{\left(3\right)}Fe\left(ỌH_2\right)\underrightarrow{\left(4\right)}FeSO_4\underrightarrow{\left(5\right)}FeCl_2\underrightarrow{\left(6\right)}Fe\left(NO_3\right)_2\)
Z/ \(Mg\left(OH\right)_2\underrightarrow{\left(1\right)}MgO\underrightarrow{\left(2\right)}MgSO_4\underrightarrow{\left(3\right)}MgCl_2\underrightarrow{\left(4\right)}Mg\left(OH\right)_2\underrightarrow{\left(5\right)}MgCl_2\underrightarrow{\left(6\right)}Mg\left(NO_3\right)_2\)
X/\(Al\left(OH\right)_3\underrightarrow{\left(1\right)}Al_2O_3\underrightarrow{\left(2\right)}AlCl_3\underrightarrow{\left(3\right)}Al\underrightarrow{\left(4\right)}Al_2\left(SO_4\right)_3\)
g) 1. Fe2O3 + 6HCl → 2FeCl3 + 3H2O
2. FeCl3 + 3NaOH → 3NaCl + Fe(OH)3↓
3. 2Fe(OH)3 \(\underrightarrow{to}\) Fe2O3 + 3H2O
4. Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
h) 1. ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓
2. Zn(OH)2 + 2HCl → ZnCl2 + 2H2O
3. ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓
4. NaCl + AgNO3 → NaNO3 + AgCl↓
m) 1. CuO + 2HCl → CuCl2 + H2O
2. CuCl2 + 2NaOH → 2NaCl + Cu(OH)2↓
3. Cu(OH)2 \(\underrightarrow{to}\) CuO + H2O
4. CuO + H2SO4 → CuSO4 + H2O
Viết các phương trình hoá học theo sơ đồ chuyển hoá sau:
CuO \(\underrightarrow{\left(1\right)}\) CuSO4 \(\underrightarrow{\left(2\right)}\) CuCl2 \(\underrightarrow{\left(3\right)}\) Cu(OH)2
\(\left(1\right)CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ \left(2\right)CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow\left(trắng\right)+CuCl_2\\ \left(3\right)CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow\left(xanh.lam\right)+2KCl\)
hoàn thành pthh
A \(\underrightarrow{+HCL}\) B \(\underrightarrow{+NaOH}\) C \(\:\underrightarrow{t}\) D \(\underrightarrow{+CO,t}\) Cu
help me......
A: CuCO3
B: CuCl2
C: Cu(OH)2
D: CuO
PTHH:
CuCO3 + HCl ===> CuCl2 + CO2 + H2O
CuCl2 + 2NaOH ===> Cu(OH)2 + 2NaCl
Cu(OH)2 =(nhiệt)=> CuO + H2O
CuO + CO =(nhiệt)=> Cu + CO2
Hoàn thành sơ đồ phản ứng sau:
NH3 \(\underrightarrow{\left(1\right)}\) NO \(\underrightarrow{\left(2\right)}\) NO2 \(\underrightarrow{\left(3\right)}\) HNO3 \(\underrightarrow{\left(4\right)}\) Cu(NO3)2 \(\underrightarrow{\left(5\right)}\) Cu(OH)2 \(\underrightarrow{\left(6\right)}\) CuO
\(4NH_3+5O_2-t^0-_{xt}>6H_2O+4NO\)
\(2NO+O_2->2NO_2\)
\(3NO_2+H_2O->2HNO_3+NO\)
\(Cu\left(OH\right)_2+2HNO_3->Cu\left(NO_3\right)_2+2H_2O\)
\(Cu\left(NO_3\right)_2+2NaOH->Cu\left(OH\right)_2+2NaNO_3\)
\(Cu\left(OH\right)_2-t^o>CuO+H_2O\)
1. Lập PTHH cho chuỗi phản ứng sau:
a. Al2O3 \(\underrightarrow{\left(1\right)}\) Al \(\underrightarrow{\left(2\right)}\) AlCl3 \(\underrightarrow{\left(3\right)}\) NaCl \(\underrightarrow{\left(4\right)}\) NaOH
b. CuSO4 \(\underrightarrow{\left(1\right)}\) Cu(OH)2 \(\underrightarrow{\left(2\right)}\) CuO \(\underrightarrow{\left(3\right)}\) Cu \(\underrightarrow{\left(4\right)}\) CuCl2
b;
CuSO4 + 2NaOH -> Cu(OH)2 + Na2SO4
Cu(OH)2 -> CuO + H2O
CuO + CO -> Cu + CO2
Cu + 2FeCl3 -> 2FeCl2 + CuCl2
a)
2Al2O3 \(\underrightarrow{đp}\) 4Al + 3O2
2Al + 3Cl2 \(\rightarrow\) 2AlCl3
AlCl3 + 3NaOH \(\rightarrow\) Al(OH)3 + 3NaCl
2NaCl + 2H2O \(\rightarrow\) 2NaOH + Cl2 + H2
a;
2Al2O3 \(\xrightarrow[criolit]{đpnc}\)4Al + 3O2
2Al + 6HCl -> 2AlCl3 + 3H2
AlCl3 + 3NaOH -> Al(OH)3 + 3NaCl
2NaCl + 2H2O \(\underrightarrow{đpdd}\) 2NaOH + H2 + Cl2
Viết các phương trình phản ứng sau:
\(Cu\underrightarrow{1}CuO\underrightarrow{2}CuCl_2\xrightarrow[4]{3}Cu\left(OH\right)_2\xrightarrow[6]{5}Cu\left(NO_3\right)_2\underrightarrow{7}Fe\left(NO_3\right)_2\xrightarrow[10]{9}Fe\left(OH\right)_2\)
\(\left(1\right)Cu+\dfrac{1}{2}O_2\xrightarrow[]{t^0}CuO\\ \left(2\right)CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(\left(3\right)CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\\ \left(4\right)Cu\left(OH\right)_2+2HNO_3\rightarrow Cu\left(NO_3\right)_2+2H_2O\\ \left(5\right)Cu\left(NO_3\right)_2+Fe\rightarrow Fe\left(NO_3\right)_2+Cu\\ \left(6\right)Fe\left(NO_3\right)_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaNO_3\)