CO2 + Ca(OH)2 => CaCO3 + H2O (1)
nCaCO3 = m/M = 25/100 = 0.25 (mol)
nCa(OH)2 = CM.V = 0.3 (mol) > 0.25
Có 2 thường hợp:
TH1: ==> 2 pứ xảy ra
2CO2 + Ca(OH)2 => Ca(HCO3)2 (2)
Theo pt ==> nCO2 (1) =nCaCO3 = 0.25 (mol)
nCO2 (2) = (0.3 - 0.25)x2 = 0.1 (mol)
nCO2 = 0.25 + 0.1 = 0.35 (mol) => V = 22.4xn = 22.4 x 0.35 = 7.84 (l)
TH2: Chỉ có 1 pứ trung hòa xảy ra: Ca(OH)2 dư
nCO2 = 0.25 (mol) => V = 22.4 x 0.25 = 5.6 (l)
\(n_{Ca\left(OH\right)_2}=0,3.1=0.03\left(mol\right)\)
\(n_{CaCO3}=\frac{25}{100}=0.25\left(mol\right)\)
TH1: Chỉ tạo ra muối CaCO3
\(n_{CO2}=n_{CaCO3}=0,25\left(mol\right)\)
\(V_{CO_2}=0,25.22,4=5.6\left(l\right)\)
TH2: Tạo ra cả hai muối
\(Ca\left(OH\right)_2+CO_{2_{ }}\rightarrow CaCO3+H_2O\)
\(Ca\left(OH\right)_2+2CO2\rightarrow Ca\left(HCO3\right)_2\)
\(n_{Ca\left(OH\right)_2\left(1\right)}=n_{CO_2\left(1\right)}=n_{CaCO3}=0,25\left(mol\right)\)
\(n_{CO_2}=2n_{Ca\left(OH\right)_2\left(2\right)}=\left(0,3-0,25\right)=0,1\left(mol\right)\)
\(\sum n_{CO_2}=0,25+0,1=0,35\left(mol\right)\)
\(V_{CO_2}=0,35.22,4=7,84\left(l\right)\)