a)
$MgCO_3 + 2HCl \to MgCl_2 + CO_2 + H_2O$
$n_{MgCO_3} = \dfrac{8,4}{84} = 0,1(mol)$
$n_{HCl} =2 n_{MgCO_3} = 0,2(mol)$
$C\%_{HCl} = \dfrac{0,2.36,5}{146}.100\% =5\%$
b)
Theo PTHH :
$n_{MgCl_2} = n_{CO_2}= n_{MgCO_3} = 0,1(mol)$
$m_{dd\ sau\ pư} = 8,4 + 146 - 0,1.44 = 150(gam)$
$C\%_{MgCl_2} = \dfrac{0,1.95}{150}.100\% = 6,33\%$


