$n_{CO_2}=\dfrac{3,36}{22,4}=0,15 mol$
$n_{Ba(OH)_2}=\dfrac{200.17,1\%}{171}=0,2 mol$
$\dfrac{2n_{Ba(OH)_2}}{n_{CO_2}}=2,6\to$ chỉ tạo muối trung hoà, dư kiềm.
$Ba(OH)_2+CO_2\to BaCO_3+H_2O$
$n_{Ba(OH)_2\text{pứ}}=n_{BaCO_3}=n_{CO_2}=0,15mol$
$\Rightarrow m_B=0,15.197=29,55g$
$m_{dd\text{spứ}}=0,15.44+200-29,55=177,05g$
$n_{Ba(OH)_2\text{dư}}=0,2-0,15=0,05 mol$
$\Rightarrow C\%A=\dfrac{0,05.171.100}{177,05}=4,83\%$