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a,Khi cho Cu(OH)2 tác dụng H2SO4 ta có Pthh:
Cu(OH)2+H2SO4\(\rightarrow\)CuSO4 +2H2O(1)
theo đề bài và pthh(1) ta có:nCu(OH)2=9,8:98=0,1 (mol)
nCu(OH)2=nH2SO4=nCuSO4=0,1(mol)
mCuSO4=0,1\(\times\)160=16(gam),V H2SO4=0,1\(\times\)22,4=2,24(l)
vậy khối lượng muối thu được là 16(gam),số lít H2SO4tham gia pư là 2,24(l)
b,khi cho KOH tác dụng với dd H2SO4 PTHH:
2KOH+H2SO4\(\rightarrow\)K2SO4+2H2O(2)
theo pthh(2) và đề bài ta có:nH2SO4=\(\dfrac{1}{2}\)nKOH\(\rightarrow\)nKOH=0,2(mol)
mKOH=0,2\(\times\)56=11,2(g),m dd KOH=11,2\(\div\)(16,8\(\div\)100)=\(\dfrac{200}{3}\) (g)
v dd KOH=\(\dfrac{200}{3}\)\(\div\)1,045=63,79(ml)
vậy v dd KOH tham gia pư là 63,79(ml)
Cu(OH)2 + H2SO4 -> CuSO4 + 2H2O (1)
nCu(OH)2=0,1(mol)
Theo PTHH 1 ta có:
nCu(OH)2=nH2SO4=nCuSO4=0,1(mol)
mCuSO4=160.0,1=16(g)
Vdd H2SO4=\(\dfrac{0,1}{1}=0,1\left(lít\right)\)
b;
2KOH + H2SO4 -> K2SO4 + 2H2O (2)
Theo PTHH 2 ta có:
nKOH=2nH2SO4=0,2(mol)
mKOH=56.0,2=11,2(g)
mdd KOH=11,2:16,8%=\(\dfrac{200}{3}\left(g\right)\)
Vdd KOH=\(\dfrac{200}{3}:1,045=63,8\left(ml\right)\)
Câu 1: Hãy viết các PTHH theo các sơ đồ phản ứng sau đây:
1) AgNO3 + HCl ---> AgCl↓+HNO3
2) Cu + H2SO4đnóng ---> CuSO4+SO2↑+H2O
3) BaCO3 + H2SO4 ---> BaSO4+CO2+H2O
4) 2NaOH + CuSO4 ---> Na2SO4+Cu(OH)2↓
5) Al(OH)3
6) K2CO3 + 2HCl --->2KCl + CO2↑+H2O
7) Ba(NO3)2 + Na2SO4 ---> NaNO3 + BaSO4↓
8) CuSO4 + 2KOH ---> K2SO4 + Cu(OH)2↓
9) AgNO3 + HCl ---> KNO3 + AgCl↓
Câu 2: Viết PTHH thực hiện chuỗi biến hóa sau:
a. Al2O3 ---> Al ---> Al(NO3)3 ---> Al(OH)3 ---> Al2O3 ---> Al2(SO4)3 ---> AlCl3 ---> Al ---> Cu
\(2Al_2O_3--dpnc->4Al+3O_2\)
\(Al+4HNO_3-->Al\left(NO_3\right)_3+NO\uparrow+2H_2O\)
\(Al\left(NO_3\right)_3+3NaOH-->Al\left(OH\right)_3\downarrow+3NaNO_3\)
\(2Al\left(OH\right)_3-to->Al_2O_3+3H_2O\)
\(Al_2O_3+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2O\)
\(Al_2\left(SO_4\right)_3+3BaCl_2-->2AlCl_3+3BaSO_4\downarrow\)
\(3Mg+2AlCl_3-->3MgCl_2+2Al\)
\(3CuCl_2+2Al-->2AlCl_3+3Cu\)
b. Fe ---> FeCl3 ---> Fe(OH)3 ---> Fe2O3 ---> Fe ---> FeCl2 ---> Fe(NO3)2 ---> FeCO3 ---> FeSO4.
\(2Fe+3Cl_2--to->2FeCl_3\)
\(FeCl_3+3NaOH-->Fe\left(OH\right)_3\downarrow+3NaCl\)
\(2Fe\left(OH\right)_3-to->Fe_2O_3+3H_2O\)
\(Fe_2O_3+3CO-to->2Fe+3CO_2\uparrow\)
\(Fe+2HCl-->FeCl_2+H_2\uparrow\)
\(FeCl_2+2AgNO_3-->Fe\left(NO_3\right)_2+AgCl\downarrow\)
\(Fe\left(NO_3\right)_2+Na_2CO_3--->FeCO_3+2NaNO_3\)
\(FeCO_3+H_2SO_4-->FeSO_4+CO_2\uparrow+H_2O\)
c. Mg ---> MgO ---> MgCl2 ---> Mg(OH)2 ---> MgSO4 ---> MgCl2 ---> Mg(NO3)2 ---> MgCO3
\(2Mg+O_2--to->MgO\)
\(MgO+2HCl-->MgCl_2+H_2O\)
\(MgCl_2+2NaOH-->Mg\left(OH\right)_2\downarrow+2NaCl\)
\(Mg\left(OH\right)_2+H_2SO_4-->MgSO_4+2H_2O\)
\(MgSO_4+BaCl_2-->MgCl_2+BaSO_4\downarrow\)
\(MgCl_2+2AgNO_3-->Mg\left(NO_3\right)_2+2AgCl\downarrow\)
\(Mg\left(NO_3\right)_2+Na_2CO_3-->MgCO_3\downarrow+2NaNO_3\)
d. Cu(OH)2 ---> CuO ---> CuSO4 ---> CuCl2 ---> Cu(NO3)2 ---> Cu ---> CuO.
\(Cu\left(OH\right)_2-->CuO+H_2O\)
\(CuO+H_2SO_4-->CuSO_4+H_2O\)
\(CuSO_4+BaCl_2-->CuCl_2+BaSO_4\downarrow\)
\(CuCl_2+2AgNO_3-->Cu\left(NO_3\right)_2+2AgCl\downarrow\)
\(Fe+Cu\left(NO_3\right)_2-->Fe\left(NO_3\right)_2+Cu\)
\(2Cu+O_2-->2CuO\)
B2/
nH2SO4= 196/98=2 mol
nNaOH = 60/40=1.5 mol
2NaOH + H2SO4 --> Na2SO4 + H2O
Bđ: 1.5________2
Pư: 1.5________0.75
Kt: 0__________1.25
2KOH + H2SO4 --> K2SO4 + H2O
2.5_______1.25
mKOH = 2.5*56=140g
mdd KOH = 140*100/40=350g
a)
NaOH + H2SO4 ---> Na2SO4 + H2O
KOH + H2SO4 ---> K2SO4 + H2O
b) nKOH = \(n_{H_2SO_4}\) - nNaOH = 2 -1,5 = 0,5( mol)
➞ mKOH = 28 g
➞ mddKOH = 70g
2Cu + O2------>2CuO (có nhiệt độ )
CuO + H2SO4(đặc nóng ) ----> CuSO4 + H2
H2 + KOH -----> K + H2O( có nhiệt độ )
1. SO2 + H2O ---> H2SO3
Na2O + H2O ---> 2NaOH
2. CuO + 2HCl ---> CuCl2 + H2O
Na2O + 2HCl ---> 2NaCl + H2O
3. CuO + H2SO4 ---> CuSO4 + H2O
Na2O + H2SO4 ---> Na2SO4 + H2O
4. 2NaOH + SO2 ---> Na2SO3 + H2O
a) H2SO4 + 2KOH → K2SO4 + 2H2O
b) \(m_{H_2SO_4}=200\times9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,4\times56=22,4\left(g\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{22,4}{8\%}=280\left(g\right)\)
a) PTHH: 3KOH + FeCl3 \(\rightarrow\) 3KCl + Fe(OH)3\(\downarrow\)(1)
2Fe(OH)3 \(\underrightarrow{t^o}\) Fe2O3 + 3H2O (2)
b) mKOH = \(\frac{300.5,6}{1000}\) = 1,68(g)
=> nKOH = \(\frac{1,68}{56}=0,03\left(mol\right)\)
Theo PT (1) : n\(Fe\left(OH\right)_3\) = \(\frac{1}{3}n_{KOH}\) = \(\frac{1}{3}.0,03=0,01\left(mol\right)\)
Theo PT(2): n\(Fe_2O_3\) = \(\frac{1}{2}n_{Fe\left(OH\right)_3}\) = \(\frac{1}{2}.0,01=0,005\left(mol\right)\)
=> m\(Fe_2O_3\) = 0,005.160 = 0,8 (g) = mB
\(\text{a, 3KOH + F e C l 3 → 3KCl + F e ( O H ) 3 (1)}\)
\(\text{ 2 F e ( O H ) 3 → F e 2 O 3 +3 H 2 O (2)}\)
Ta có :
\(m_{KOH}=\frac{5,6.300}{100}\text{= 16.8 g}\)
\(\Rightarrow n_{KOH}\approx\text{ 0,285 mol}\)
Theo pthh (1) : \(n_{Fe\left(OH\right)3}=\frac{1}{3}n_{KOH}\text{= 0,095 mol}\)
Theo pthh (2) : \(n_{Fe2O3}=\frac{1}{2}n_{Fe\left(OH\right)3}=\text{0,0475 mol}\)
\(\text{⇒ m F e 2 O 3 = 7,6g}\)
Theo đề bài ta có : ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩VddH2O4=601,2=50(ml)nNaOH=20.20100.40=0,1(mol){VddH2O4=601,2=50(ml)nNaOH=20.20100.40=0,1(mol)
nFe = 1,68/56 = 0,03 mol
a) Ta có PTHH :
2NaOH + H2SO4 -> Na2SO4 + 2H2O
0,1mol......0,05mol
=> CMH2SO4 = 0,05/0,05=1(M)
PTHH: \(Na_2O\left(0,5\right)+H_2O\rightarrow2NaOH\left(1\right)\)
Phần 1: \(6NaOH\left(1\right)+Fe_2\left(SO_4\right)_3\left(\dfrac{1}{6}\right)\rightarrow3Na_2SO_4\left(0,5\right)+2Fe\left(OH\right)_3\)
\(\Rightarrow V=x=\dfrac{\dfrac{1}{6}}{0,5}\approx0,3M\)
\(C_{MddNa_2SO_4}=\dfrac{0,5}{0,5}=1M\)
Phần 2: \(2NaOH\left(1\right)+H_2SO_4\left(0,5\right)\rightarrow Na_2SO_4\left(0,5\right)+2H_2O\)
\(m_{H_2SO_4}=0,5.98=49g\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{100.49}{20}=245g\Rightarrow V_{ddH_2SO_4}=\dfrac{245}{1,14}\approx214,9ml\)
\(m_{Na_2SO_4}=0,5.142=71g\)
\(n_{H_2SO_4}=0,8.0,25=0,2mol\)
H2SO4+2KOH\(\rightarrow\)K2SO4+H2O
\(n_{KOH}=2n_{H_2SO_4}=0,4mol\)
mKOH=0,4.56=22,4g
C%\(=\dfrac{22,4.100}{500}=4,48\%\)
CO2+KOH\(\rightarrow\)KHCO3
\(n_{CO_2}=n_{KOH}=0,4mol\)
\(V_{CO_2}=0,4.22,4=8,96l\)