PTHH: MgCO3 + 2HCl \(\rightarrow\) MgCl2 + H2O + CO2\(\uparrow\)(1)
CO2 + Ca(OH)2 \(\rightarrow\) CaCO3 \(\downarrow\) + H2O (2)
n\(MgCO_3\) = \(\frac{9,4}{84}=\frac{47}{420}\left(mol\right)\)
Theo PT(1): n\(CO_2\) = n\(MgCO_3\) = \(\frac{47}{420}\left(mol\right)\) = n\(CO_2\)(2)
Theo PT(2): n\(CaCO_3\) = n\(CO_2\) = \(\frac{47}{420}\left(mol\right)\)
=> m\(CaCO_3\) = \(\frac{47}{420}.100\) = 11,19 (g)
Bạn sửa đề lại là : 8.4g MgCO3 nhé
nMgCO3 = 8.4/84=0.1 mol
MgCO3 + 2HCl --> MgCl2 + CO + H2O
0.1______________________0.1
Ca(OH)2 + CO2 --> CaCO3 + H2O
0.1_______________0.1
mCaCO3 = 0.1*100=10g