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PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)
a+b) Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,2\left(mol\right)=n_{KOH}\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2\cdot36,5}{300}\cdot100\%\approx2,43\%\\C_{M_{KOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
c) PTHH: \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Theo các PTHH: \(n_{CuO\left(lý.thuyết\right)}=n_{Cu\left(OH\right)_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow n_{CuO}=0,1\cdot95\%=0,095\left(mol\right)\) \(\Rightarrow m_{CuO}=0,095\cdot80=7,6\left(g\right)\)
a, \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)
b, \(n_{KOH}=0,12.0,4=0,048\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=0,024\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,024}{0,08}=0,3\left(M\right)\)
c, \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,024\left(mol\right)\)
\(\Rightarrow C_{M_{K_2SO_4}}=\dfrac{0,024}{0,08+0,12}=0,12\left(M\right)\)
\(n_{H^+}=0,07mol=n_{OH^-}\)=>\(v=\dfrac{0,07}{0,2+0,1.2}=0,175l\)
200ml = 0,2l
\(n_{Ba\left(OH\right)2}=0,5.0,2=0,1\left(mol\right)\)
Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)
1 2 1 2
0,1 0,2 0,1
a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)
b) \(n_{BaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{BaCl2}=0,1.208=20,8\left(g\right)\)
c) \(V_{ddspu}=0,2+0,2=0,4\left(l\right)\)
\(C_{M_{BaCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)
Chúc bạn học tốt
PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Ta có: \(n_{Ba\left(OH\right)_2}=0,2\cdot0,5=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\\m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\\C_{M_{BaCl_2}}=\dfrac{0,1}{0,2+0,2}=0,25\left(M\right)\end{matrix}\right.\)
Bài 1 :
nNaOH = 0,6 (mol)
NaOH + HCl -> NaCl + H2O
0,6...........0,6........0,6 (mol)
mdd HCl = \(\frac{0,6.36,5}{7,3\%}=300\left(g\right)\)
\(C\%_{NaCl}=\frac{0,6.58,5}{300+200}.100\%=7,02\%\)
a)Phương Trình Hóa Học :
HCl + NaOH → NaCl + H2O
nHCl = 0,04 (mol) = nNaOH = nNaCl
=>VddNaOH = 0,04/0,1 = 0,4 (l) = 400 (ml)
Vdd = VddNaOH + VddHCl = 0,6 (l)
=>CM ≈ 0,067 (M)
b) 2HCl + Ca(OH)2 → CaCl2 + 2H2O
nCa(OH)2 = nCaCl2 = (1/2)nHCl = 0,02 (mol)
(Nồng độ phần trăm = 25% ????)
mCa(OH)2 = 1,48 (g)
=>mdd(Ca(OH)2) = 5,92 (g)
mddHCl = 220 (g)
=>mdd = 225,92 (g)
mCaCl2 = 2,22 (g)
=>%mCaCl2 ≈ 0,98%
a, \(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\)
b, \(n_{HCl}=0,06.0,1=0,006\left(mol\right)\)
Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,003\left(mol\right)\)
\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,003}{0,2}=0,015\left(l\right)=15\left(ml\right)\)
c, \(n_{BaCl_2}=\dfrac{1}{2}n_{Ba\left(OH\right)_2}=0,003\left(mol\right)\Rightarrow C_{M_{BaCl_2}}=\dfrac{0,003}{0,06+0,015}=0,04\left(M\right)\)
\(a/2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\\ b/n_{HCl}=0,06.0,1=0,006mol\\ n_{Ba\left(OH\right)_2}=n_{BaCl_2}=0,006:2=0,003mol\\ V_{Ba\left(OH\right)_2}=\dfrac{0,003}{0,2}=0,015l\\ c/C_{M_{BaCl_2}}=\dfrac{0,003}{0,06+0,015}=0,04M\)