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Bài 1:
\(2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Na}=n_{Na_2O}=0,2.2=0,4\left(mol\right)\\ a.m_{Na}=0,4.23=9,2\left(g\right)\\ b.C_{MddA}=\dfrac{0,4}{0,5}=0,8\left(M\right)\\ C\%_{ddA}=\dfrac{0,4.40}{500.1,2}.100\approx2,667\%\)
a, \(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)
PTHH: K2O + H2O → 2KOH
Mol: 0,05 0,1
b) \(C_{M_{ddKOH}}=\dfrac{0,1}{0,02}=5M\)
c)
PTHH: KOH + HCl → KCl + H2O
Mol: 0,1 0,1 0,1
\(m_{ddHCl}=\dfrac{0,1.36,5.100}{20}=18,25\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{18,25}{0,9125}=103,9\left(ml\right)=0,1039\left(l\right)\)
d) \(C_{M_{ddKCl}}=\dfrac{0,1}{0,02+0,1039}=0,8071M\)
a)
$2K + 2H_2O \to 2KOH + H_2$
$BaO + H_2O \to Ba(OH)_2$
Theo PTHH :
$n_K = 2n_{H_2} = 0,2(mol)$
$\%m_K = \dfrac{0,2.39}{23,1}.100\% = 33,77\%$
$\%m_{BaO} = 100\%- 33,77\% = 66,23\%$
b)
$n_{BaO} = \dfrac{23,1 - 0,2.39}{153} = 0,1(mol)$
$m_{dd} = 23,1 + 177,1 - 0,1.2 = 200(gam)$
$C\%_{KOH} = \dfrac{0,2.56}{200}.100\% = 5,6\%$
$C\%_{Ba(OH)_2} = \dfrac{0,1.171}{200}.100\% = 8,55\%$
c)
$KOH + HCl \to KCl + H_2O$
$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} + n_{KOH} = 0,4(mol)$
$V = \dfrac{0,4}{0,5} = 0,8(lít) = 800(ml)$
a, Ta có: 27nAl + 56nFe = 22 (1)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)
b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)
1. \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,25 0,5
\(C_{M_{ddNaOH}}=\dfrac{0,5}{0,5}=1M\)
2.
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,5 0,25
\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\)
\(V_{ddH_2SO_4}=\dfrac{122,5}{1,14}=107,456\left(ml\right)\)
\(n_{Na2O}=\dfrac{m_{Na2O}}{M_{Na2O}}=0,25\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,25 mol - 0,25 mol - 0,5 mol
a) \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{V_{NaOH}}=1\left(M\right)\)
b) \(H_2SO_4+2NaOH\rightarrow Na_2SO4+2H_2O\)
0,25 mol - 0,5 mol - 0,25 mol - 0,5 mol
\(m_{ctH2SO4}=n_{H2SO4}.M_{H2SO4}=24,5\left(g\right)\)
\(C_{\%_{H2SO4}}=\dfrac{m_{ctH2SO4}}{m_{ddH2SO4}}.100\%\)
\(\Rightarrow m_{ddH2SO4}=\dfrac{m_{ctH2SO4}.100\%}{C_{\%_{H2SO4}}}=122,5\left(g\right)\)
\(D_{H2SO4}=\dfrac{m_{ddH2SO4}}{V_{H2SO4}}\Rightarrow V_{H2SO4}=\dfrac{m_{ddH2SO4}}{D_{H2SO4}}\approx107,46\left(ml\right)\)
BTKL: mD + mNaHCO3 = mCO2 + mE
mD + 179,88 = 44.0,2 + 492 => mD = 320,92
BTKL: mMg + mddHCl = mH2 + mD
=> 24 . 0,4 + mddHCl = 2 . 0,4 + 320,92 => mddHCl = 312,12
=> C%HCl = 11,69%
\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)
Gọi $n_{Na_2O} = 2a(mol) \Rightarrow n_{K_2O} = a(mol)$
$\Rightarrow 2a.62 + 94a = 21,8 \Rightarrow a = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$K_2O + H_2O \to 2KOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$n_{KOH} = 2n_{K_2O} = 0,2(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} = 0,8M$
$C_{M_{KOH}} = \dfrac{0,2}{0,5} = 0,4M$
$m_{dd} = D.V = 1,04.500 = 520(gam)$
$C\%_{NaOH} = \dfrac{0,4.40}{520}.100\% = 3,1\%$
$C\%_{KOH} = \dfrac{0,2.56}{520}.100\% = 2,15\%$