a,
$CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$
$MgCO_3+2HCl\to MgCl_2+CO_2+H_2O$
$Na_2CO_3+2HCl\to 2RCl+CO_2+H_2O$
b,
$n_{HCl}=0,2.4=0,8(mol)$
$\Rightarrow n_{CO_2}=n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,4(mol)$
BTKL:
$m=39,6+0,8.36,5-0,4.44-0,4.18$
$=44g$
c,
$n_{NaOH}=\dfrac{60.40\%}{40}=0,6(mol)$
$\dfrac{n_{NaOH}}{n_{CO_2}}=1,5$
$\Rightarrow$ tạo muối $NaHCO_3$ (x mol), $Na_2CO_3$ (y mol)
$NaOH+CO_2\to NaHCO_3$
$2NaOH+CO_2\to Na_2CO_3+H_2O$
$\Rightarrow x+2y=0,6; x+y=0,4$
$\Leftrightarrow x=y=0,2$
$\Rightarrow m_{\text{muối}}=106.0,2+84.0,2=38g$
