a,
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\left(1\right)\)
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\left(2\right)\)
Khí A là CO2
Dung dịch B là MgCl2
Kết tủa C là Mg(OH)2.
\(n_{MgCO3}=\frac{16,8}{84}=0,2\left(mol\right)\)
Theo PTHH 1:
\(n_{MgCl2}=n_{MgCO3}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgCl2}=0,2.95=19\left(g\right)\)
\(m_{dd\left(spu\right)}=m_{MgCO3}+m_{dd\left(HCl\right)}-m_{CO2}=16,8+200-\left(0,2.44\right)=208\left(g\right)\)
\(\Rightarrow C\%_{dd\left(MgCl2\right)}=\frac{19}{208}.100\%=9,13\%\)
b,\(n_{Mg\left(OH\right)2}=n_{MgCl2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg\left(OH\right)2}=0,2.58=11,6\left(g\right)\)
