a) CO2+Ca(OH)2--->CaCO3+H2O
n\(_{CO2}=\frac{1,568}{22,4}=0,07\left(mol\right)\)
n\(_{Ca\left(OH\right)2}=\frac{6,4}{74}=0,086\left(mol\right)\)
=>CO2 dư..Tạo 1muối
Theo pthh
n\(_{Ca\left(OH\right)2}=n_{CO2}=0,07\left(mol\right)\)
n Ca(OH)2 dư=0,086-0,07=0,016(mol)
m\(_{Ca\left(OH\right)2}=0,016.74=1,184\left(g\right)\)
b)Theo pthh1
n\(_{_{ }CaCO3}=n_{CO2}=0,07\left(mol\right)\)
m\(_{CaCO3}=0,07.100=7\left(g\right)\)
Ta có : \(\text{n CO2= 0,07 mol}\)
\(\text{n Ca(OH)2 = 16/185 (mol)}\)
\(PTHH:CaO+Ca\left(OH\right)2\rightarrow CaCO3+H2O\)
Theo pt pứ: 1 mol CO2 pứ vs 1 mol Ca(OH)2
mà n Ca(OH)2 > n CO2
--> Ca(OH)2 dư, n Ca(OH)2 pứ = n CO2 =0,07 mol.
\(\text{m Ca(OH)2 dư = 6,4 - 0,07.74= 1,22g}\)
\(\text{b, m CaCO3 = 0,07.100=7g}\)
Sửa đề là : 1.568 (l) CO2 nhé
nCO2 = 1.568/22.4 = 0.07 mol
nCa(OH)2 = 6.4/74=16/185 mol
Ca(OH)2 + CO2 --> CaCO3 + H2O
Bđ: 16/185_____0.07
Pư: 0.07________0.07____0.07
Kt: 61/3700______0______0.07
=> Ca(OH)2 dư
mCa(OH)2 = 61/3700*74=1.22 g
mCaCO3 = 0.07*100=7 g
