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nCuSO4=0,5.0,4=0,2 mol
CuSO4 +2NaOH=> Cu(OH)2+Na2SO4
0,2 mol =>0,2 mol
Cu(OH)2=> CuO+H2O
0,2 mol =>0,2 mol
kết tủa A là Cu(OH)2 m=98.0,2=19,6g
cr B là CuO m=0,2.80=16g
a) \(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
PTHH: CuCl2 + 2NaOH --> Cu(OH)2 + 2NaCl
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => CuCl2 hết, NaOH dư
PTHH: CuCl2 + 2NaOH --> Cu(OH)2 + 2NaCl
0,2------>0,4-------->0,2------->0,4
Cu(OH)2 --to--> CuO + H2O
0,2-------------->0,2
=> mCuO = 0,2.80 = 16(g)
b)
\(\left\{{}\begin{matrix}m_{NaOH\left(dư\right)}=20-0,4.40=4\left(g\right)\\m_{NaCl}=0,4.58,5=23,4\left(g\right)\end{matrix}\right.\)
\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
PT: CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4
mol 0,02 0,04 ← 0,02 → 0,02
2Cu(OH)2 + O2 \(\underrightarrow{t^o}\) 2CuO + 2H2O
mol 0,02 0,01 ← 0,02 → 0,02
a) \(m_{Cu\left(OH\right)_2}=0,02.98=1,96\left(g\right)\)
b) Vdung dịch sau phản ứng = 200 + 300 = 500 (ml) = 0,5 (l)
\(C_{MNa_2SO_4}=\dfrac{0,02}{0,5}=0,04M\)
\(n_{CuSO_4}=0.2\cdot0.5=0.1\left(mol\right)\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
\(0.1.............0.2.................0.1..........0.1\)
\(C_{M_{Na_2SO_4}}=\dfrac{0.1}{0.3+0.2}=0.2\left(M\right)\)
\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)
\(0.1.............0.1\)
\(m_{CuO}=0.1\cdot80=8\left(g\right)\)
a)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow3Na_2SO_4+2Al\left(OH\right)_3\downarrow\)
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
b) B gồm Fe(OH)2, Cu(OH)2
C gồm CuO, Fe2O3
a)\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\)
\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)
b)B là các chất sau: \(Fe\left(OH\right)_2;Cu\left(OH\right)_2;Al\left(OH\right)_3\)
C là các chất sau: \(Fe_2O_3;CuO;Al_2O_3\)
a) 3NaOH+FeCl3---->Fe(OH)3+3NaCl
b) n NaOH=0,15.2=0,3(mol)
Theo pthh
n FeCl3=1/3n NaOH=0,1(mol)
m dd FeCl3=\(\frac{162,5.100.0,1}{20}=81,25\left(g\right)\)
c) Theo pthh
n Fe(OH)3=1/3n NaOH=0,1(mol)
m Fe(OH)3=0,1.107=10,7(g)
d) 2Fe(OH)3--->Fe2O3+3H2O
Theo pthh
n Fe2O3=1/2n Fe(OH)3=0,05(mol)
m Fe2O3=0,05.160=8(g)
PTPU
Mg+ 2HCl\(\rightarrow\) MgCl2+ H2\(\uparrow\) (1)
MgCl2+ 2NaOH\(\rightarrow\) Mg(OH)2\(\downarrow\)+ 2NaCl (2)
Mg(OH)2\(\xrightarrow[]{to}\) MgO+ H2O (3)
có: nMg= \(\frac{4,8}{24}\)= 0,2( mol)
theo ptpư(1) có: nHCl= 2nMg= 0,4( mol)
\(\Rightarrow\) mdd HCl= \(\frac{0,4.36,5}{20\%}\)= 73( g)
có: nMgCl2= nH2= nMg= 0,2( mol)
\(\Rightarrow\) mMgCl2= 0,2. 95= 19( g)
có: mdd sau pư= mMg+ mdd HCl- mH2
= 4,8+ 73- 0,2. 2= 77,4( g)
\(\Rightarrow\) C%MgCl2= \(\frac{19}{77,4}\). 100%= 24,55%
theo ptpư(2) có: nMg(OH)2= nMgCl2= 0,2( mol)
\(\Rightarrow\) mMg(OH)2= 0,2. 58= 11,6( g)
theo ptpư(3) có: nMgO= nMg(OH)2= 0,2( mol)
\(\Rightarrow\) mMgO= 0,2. 40=8( g)