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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) : \(n_{NaOH}=n_{NaCl}=0,5\left(mol\right)\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)
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: \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\)
\(Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\)
\(m_{CuO}=0,03.80=2,4g\)
a) PTHH: \(3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,5\cdot0,1=0,05\left(mol\right)\\n_{FeCl_3}=0,2\cdot0,2=0,04\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,05}{3}< \dfrac{0,04}{1}\) \(\Rightarrow\) NaOH p/ứ hết, FeCl3 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,05\left(mol\right)\\n_{FeCl_3\left(dư\right)}=\dfrac{7}{300}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,05\cdot58,5=2,925\left(g\right)\\m_{FeCl_3\left(dư\right)}=\dfrac{7}{300}\cdot162,5\approx3,8\left(g\right)\end{matrix}\right.\)
c) PTHH: \(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)
Ta có: \(n_{Fe\left(OH\right)_3}=\dfrac{1}{60}\left(mol\right)\) \(\Rightarrow n_{H_2O}=\dfrac{1}{40}\left(mol\right)\) \(\Rightarrow m_{H_2O}=\dfrac{1}{40}\cdot18=0,45\left(g\right)\)
a, \(CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
b, \(n_{CuCl_2}=\dfrac{20,25}{135}=0,15\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{CuCl_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,3}{0,3}=1\left(M\right)\)
c, \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)
