\(n_{K_2SO_4}=1\cdot0,2=0,2\left(mol\right)\\ n_{BaCl_2}=2\cdot0,15=0,3\left(mol\right)\\ PTHH:K_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2KCl\\ \text{Vì }\dfrac{n_{K_2SO_4}}{1}< \dfrac{n_{BaCl_2}}{1}\text{ nên sau p/ứ }BaCl_2\text{ dư}\\ \Rightarrow n_{BaSO_4}=n_{K_2SO_4}=0,2\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{BaSO_4}=0,2\cdot233=46,6\left(g\right)\)