CaO + 2HCl → CaCl2 + H2O (1)
CaCO3 + 2HCl → CaCl2 + CO2↑ + H2O (2)
\(n_{CO_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT2: \(n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,2\times100=20\left(g\right)\)
\(\Sigma n_{CaCl_2}=\frac{66,6}{111}=0,6\left(mol\right)\)
Theo PT2: \(n_{CaCl_2}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow n_{CaCl_2}\left(1\right)=0,6-0,2=0,4\left(mol\right)\)
Theo PT1: \(n_{CaO}=n_{CaCl_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{CaO}=0,4\times56=22,4\left(g\right)\)
Theo PT1: \(n_{HCl}=2n_{CaO}=2\times0,4=0,8\left(mol\right)\)
Theo Pt2: \(n_{HCl}=2n_{CaCO_3}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,8+0,4=1,2\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=1,2\times36,5=43,8\left(g\right)\)
nCO2 = 4.48/22.4 = 0.2 mol
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
0.2_______0.4______0.2_____0.2
nCaCl2 = 66.6/ 111=0.6 mol
=> nCaCl2 (còn lại ) = 0.6 - 0.2 = 0.4 mol
CaO + 2HCl --> CaCl2 + H2O
0.4____0.8______0.4
mCaCO3 = 0.2*100=20 g
mCaO = 0.4*56=22.4g
nHCl = 0.4 + 0.8 = 1.2 mol
mHCl = 1.2*36.5=43.8g
