$K_2CO_3 + 2HCl => 2KCl + H_2O + CO_2$$n_{K_2CO_3} = 69/138 = 0,5$ mol
$n_{CO_2} = n_{K_2CO_3} = 0,5$ mol => $m_{CO_2} = 0,5.44 = 22$ gam
=> Bình $A$ tăng : $69-22 = 47$ gam
$CaCO_3 + 2HCl => CaCl_2 + H_2O + CO_2$
$n_{CO_2} = n_{CaCO_3} = m/100 => m_{CO_2} = 44m/100 = 11m/25$ gam
=> Bình $B$ tăng : $m - 11m/25 = 14m/25$ gam
do cân thăng bằng => $14m/25 = 47 => m = 83,93$ gam