$n_{CO_2} = \dfrac{7{,}437}{22{,}4} = 0{,}3329 , \text{mol}$
$n_{Ba(OH)_2} = 0{,}2 \times 0{,}5 = 0{,}1 , \text{mol}$
Phản ứng: $Ba(OH)_2 + 2CO_2 \rightarrow Ba(HCO_3)_2$
$n_{Ba(HCO_3)_2} = 0{,}1 , \text{mol}$
$M_{Ba(HCO_3)_2} = 137 + 2(1 + 12 + 3 \times 16) = 259 , \text{g/mol}$
$m = 0{,}1 \times 259 = \boxed{25{,}9 , \text{gam}}$