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\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)
a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)
b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)
c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)
\(a.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ 0,25.......0,25............0,25..........0,25\left(mol\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\ b.m_{\downarrow}=m_{CaCO_3}=100.0,25=25\left(g\right)\)
a, \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
b, \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,25\left(mol\right)\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)
c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\approx121,67\left(g\right)\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\\ a)CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(OH\right)_2+H_2O\)
0,25 0,25 0,25
\(b)C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25M\\ c)2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\\ n_{HCl}=2n_{Ca\left(OH\right)_2}=2.0,25=0,5mol\\ m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\cdot100\%\approx121,67g\)
PTHH: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Đổi 1 bar = 0,98692 atm
a+b) \(n_{CO_2}=\dfrac{PV}{RT}=\dfrac{0,98692\cdot6,2}{0,082\cdot\left(25+273\right)}=0,25\left(mol\right)=n_{CaCO_3}=n_{Ca\left(OH\right)_2}\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\m_{CaCO_3}=0,25\cdot100=25\left(g\right)\end{matrix}\right.\)
c) PTHH: \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
Theo PTHH: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{20\%}=91,25\left(g\right)\)
a/ PTHH: CO2 + Ca(OH)2 ===> CaCO3 + H2O
b/ nCO2 = 2,24 / 22,4 = 0,1 mol
=> nCa(OH)2 = nCO2 = 0,1 mol
=>CM[Ca(OH)2] = 0,1 / 0,2 = 0,5M
c/ nCaCO3 = nCO2 = 0,1 mol
=> mCaCO3 = 0,1 x 100 = 10 gam
\(a.n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ 0,1...........0,1.............0,1..........0,1\left(mol\right)\\ b.m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\\ c.C_{MddBa\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
\(n_{SO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
TH1 : Tạo ra muối trung hòa
\(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)
\(0.25...........0.25\)
\(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0.25}{0.1}=2.5\left(M\right)\)
TH2 : Tạo ra muối axit
\(Ca\left(OH\right)_2+2SO_2\rightarrow Ca\left(HSO_3\right)_2\)
\(0.125............0.25\)
\(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0.125}{0.1}=1.25\left(M\right)\)
a, \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: CO2 + Ca(OH)2 → CaCO3 ↓ + H2O
Mol: 0,25 0,25 0,25
\(C_{M_{ddCa\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5M\)
b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)
c,
PTHH: Ca(OH)2 + 2HCl → CaCl2 + 2H2O
Mol: 0,25 0,5
\(m_{ddHCl}=\dfrac{0,5.36,5.100}{20}=91,25\left(g\right)\)