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Đáp án:

$n_{Ba}=\dfrac{41,1}{137}=0,3(mol)$

$n_{H_2SO_4}=\dfrac{110.9,8\%}{98}=0,11(mol)$ 

$Ba+H_2SO_4\to BaSO_4+H_2$  

$0,3>0,11\Rightarrow Ba$ dư, $H_2SO_4$ hết

$n_{Ba\text{pứ}}=n_{H_2SO_4}=n_{BaSO_4}= n_{H_2}=0,11(mol)$ 

$n_{Ba\text{dư}}=0,3-0,11=0,19(mol)$  

$Ba+2H_2O\to Ba(OH)_2+H_2$ 

$n_{Ba(OH)_2}=n_{H_2}=n_{Ba\text{dư}}=0,19(mol)$ 

$\Rightarrow \sum n_{H_2}=0,11+0,19=0,3(mol)$ 

$m_{\text{dd sau}}=41,1+110-0,11.233-0,3.2=124,87g$ 

Vậy $C\%_{Ba(OH)_2}=\dfrac{0,19.171.100\%}{124,87}=26,02\%$