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\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
a) \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\left(1\right)\)
\(Cu\left(OH\right)_2\xrightarrow[t^o]{}CuO+H_2O\left(2\right)\)
b) \(Pt\left(1\right):n_{Cu\left(OH\right)2}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(Pt\left(2\right):n_{Cu\left(OH\right)2}=n_{CuO}=0,25\left(mol\right)\Rightarrow m_{Cu}=0,25.64=16\left(g\right)\)
c) Pt(1) : \(n_{NaOH}=n_{NaCl}=0,5\left(mol\right)\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)
FeCl2+ 2NaOH --> Fe(OH)2 + 2NaCl (1)
4Fe(OH)2 +O2 --to-> 2Fe2O3 + 4H2O (2)
nFeCl2=0,2(mol)
nNaOH=0,5(mol)
Lập tỉ lệ :
\(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)
=> FeCl2 hết ,NaOH dư
Theo (1,2) : nFe2O3=1/2nFeCl2=0,1(mol)
=> x=16(g)
b) VNaOH=\(\dfrac{m}{D}=\dfrac{200}{1,12}\approx178,6\left(ml\right)\)\(\approx\)0,1786(l)
Theo (1) : nNaOH(PƯ)=2nFeCl2=0,4(mol)
=>nNaOH dư=0,1(mol)
nNaCl=2nFeCl2=0,4(mol)
=> CM dd NaCl\(\approx\)2,24(M)
CM dd NaOH dư\(\approx\)0,6(M)
a, \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\)
\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
b, \(n_{FeCl_3}=0,4.2=0,8\left(mol\right)\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=2,4\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{2,4}{0,4+0,2}=4\left(M\right)\)
c, \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=\dfrac{1}{2}n_{FeCl_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,4.160=64\left(g\right)\)
\(n_{FeCl3}=2.0,4=0,8\left(mol\right)\)
PTHH : \(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
0,8----------------------->0,8----------->2,4
b) \(C_{MNaCl}=\dfrac{2,4}{0,4+0,2}=4M\)
c) \(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)
0,8--------------->0,4
\(\Rightarrow a=m_{Fe2O3}=0,4.160=64\left(g\right)\)
nZnCl2 =40,8/136=0,3mol
nNaOH= 0,1.0,5=0,05mol
a)
pt : ZnCl2 + 2NaOH ------> Zn(OH)2\(\downarrow\) + 2NaCl
ncó: 0,3 0,05
n pứ: 0,025<------0,05-------->0,025-------->0,05
n dư: 0,275 0
b)
mZnCl2 dư = 0,275.136=37,4g
mNaCl=0,05.58,5=2,925g
c)
pt : Zn(OH)2 ---to--> ZnO + H2O
n pứ : 0,025------------>0,025
mZnO=0,025.81=2,025g
d)
vdd sau pứ =Vdd NaOH =0,1l
CM(ZnCl2 dư )=0,025/0,1=0,25M
CM(NaOH)=0,05/0,1= 0,5M
\(n_{CuCl_2}=0,1.0,3=0,03mol\)
PTHH: \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\)
\(Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\)
\(m_{CuO}=0,03.80=2,4g\)
CaCl2 trộn với NaOH không tạo kết tủa nha em!
thực tế thì p/ứ tạo ra Ca(OH)2 kết tủa đó a :))