a) PTHH: CaCO3 + 2HCl = CaCl2 + H2O + CO2
b) nCaCO3 = \(\dfrac{21}{100}\) = 0,21 mol
Theo PTHH nCaCO3 = \(\dfrac{n_{HCl}}{2}\) \(\Rightarrow\) nHCl = 0,21*2 = 0,42 (mol)
VHCl = \(\dfrac{0.42}{2}\) = 0,21 (lít)
c) PTHH : CO2 + Ca(OH)2 \(\Rightarrow\) CaCO3 + H2O
nCO2 = nCaCO3 = 0,21 mol
mkết tủa = mCaCO3 = 100*0,21 =21 (g)