Đáp án:
$a)Na_2CO_3+2HCl \to 2NaCl+CO_2+H_2O$
$b)m_{HCl}=250.7,3\%=18,25(g)$
$n_{HCl}=\dfrac{18,25}{36,5}=0,5(mol)$
$n_{CO_2}=\dfrac{1}{2}.0,5=0,25(mol)$
$V_{H_2}=0,25.22,4=5,6(l)$
$c) n_{Na_2CO_3}=n_{CO_2}=0,25(mol)$
$C\%_{Na_2CO_3}=\dfrac{0,25.(23.2+12+16.3)}{200\%}=13,25\%$