Câu 1:

$n_{CaCO_3}=\dfrac{12}{100}=0,12(mol)$

$n_{Ca(HCO_3)_2}=\dfrac{12,15}{162}=0,075(mol)$

$CO_2+Ca(OH)_2\to CaCO_3+H_2O$

$2CO_2+Ca(OH)_2\to Ca(HCO_3)_2$

$\Rightarrow n_{CO_2}=n_{CaCO_3}+2n_{Ca(HCO_3)_2}=0,27(mol)$

$\Rightarrow V=0,27.22,4=6,048l$

$n_{Ca(OH)_2}=0,12+0,075=0,195(mol)$

$\Rightarrow m_{dd Ca(OH)_2}=0,195.74:7,4\%=195g$

$Ca(HCO_3)_2+Ca(OH)_2\to 2CaCO_3+2H_2O$

$\Rightarrow n_{CaCO_3}=2n_{Ca(HCO_3)_2}=0,15(mol)$

$\to m_{\downarrow}=0,15.100=15g$

Câu 2: không tính được (các chất không phản ứng)

Bài 1:

$a,CO_2+Ca(OH)_2→CaCO_3+H_2O$

$CaCO_3+H_2O+CO_2→Ca(HCO_3)_2$

$n_{Ca(HCO_3)_2}=12,15/162=0,075(mol)$

$⇒n_{CaCO_3p.ư}=0,075(mol)$

$⇒n_{CaCO_3}=0,075+12/100=0,195(mol)$

$⇒m_{Ca(OH)_2}$=0,195.74=14,43(g)$

$⇒m_{ddCa(OH)_2}=14,43.100/7,4=195(g)$

$n_{CO_2}=n_{Ca(OH)_2}+n_{CaCO_3}=0,27(mol)$

$V_{CO_2}=0,27.22,4=6,048(l)$

$b,Ca(OH)_2+Ca(HCO_3)_2→2CaCO_3+2H_2O$

$n_{CaCO_3}=2n{Ca(HCO_3)_2}=0,075.2=0,15(mol)$

$⇒m_{kt}=0,15.100=15(g)$