\(\left(NH_4\right)_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2NH_3\uparrow+2H_2O\\ n_{NH_3}=0,15\left(mol\right)\\ \Rightarrow n_{\left(NH_4\right)_2SO_4}=n_{Ba\left(OH\right)_2}=n_{BaSO_4}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ m_1=m_{dd\left(NH_4\right)_2SO_4}=\dfrac{0,075.132.100}{13,2}=75\left(g\right)\\ m_2=m_{ddBa\left(OH\right)_2}=\dfrac{0,075.171.100}{25}=51,3\left(g\right)\\ m_{\downarrow}=m_{BaSO_4}=0,075.233=17,475\left(g\right)\)