a,
$m_{dd H_2SO_4}=100.1,14=114g$
$\Rightarrow n_{H_2SO_4}=\dfrac{114.20\%}{98}=0,23(mol)$
$n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1(mol)$
$BaCl_2+H_2SO_4\to BaSO_4+2HCl$
$\Rightarrow n_{BaSO_4}=0,1(mol)$
$m_{\downarrow}=0,1.233=23,3g$
b,
$m_{dd\text{spu}}=114+400-23,3=490,7g$
$n_{HCl}=0,1.2=0,2(mol)\Rightarrow C\%_{HCl}=\dfrac{0,2.36,5.100}{490,7}=1,49\%$
$n_{H_2SO_4\text{dư}}=0,23-0,1=0,13(mol)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,13.98.100}{490,7}=2,6\%$