Bài 1:

$200(ml)=0,2(l)\\ PTHH:2NaOH+SO_2\to Na_2SO_3+H_2O\\ n_{SO_2}=\dfrac{6,4}{64}=0,1(mol);n_{NaOH}=1,5.0,2=0,3(mol)\\ \text{Lập tỉ lệ: }\dfrac{n_{SO_2}}{1}<\dfrac{n_{NaOH}}{2}\Rightarrow NaOH\text{ dư}\\ \Rightarrow n_{Na_2SO_3}=n_{SO_2}=0,1(mol)\\ n_{NaOH(dư)}=0,3-0,1.2=0,1(mol)\\ \Rightarrow C_{M_{Na_2SO_3}}=\dfrac{0,1}{0,2}=0,5M\\ C_{M_{NaOH(dư)}}=\dfrac{0,1}{0,2}=0,5M$

Bài 2:

$a,PTHH:Ba+2H_2O\to Ba(OH)_2+H_2\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Ba}=n_{H_2}=0,15(mol)\\ \Rightarrow m_{Ba}=0,15.137=20,55(g)\\ c,500(ml)=0,5(l)\\ n_{Ba(OH)_2}=n_{H_2}=0,15(mol)\\ \Rightarrow C_{M_{Ba(OH)_2}}=\dfrac{0,15}{0,5}=0,3M\\ d,300(ml)=0,3(l)\\ PTHH:Ba(OH)_2+Na_2SO_4\to BaSO_4\downarrow+2NaOH\\ n_{Na_2SO_4}=0,3.0,3=0,09(mol)\\ \text{Lập tỉ lệ: }\dfrac{0,15}{1}>\dfrac{0,09}{1}\Rightarrow Ba(OH)_2\text{ dư}\\ \Rightarrow n_{BaSO_4}=n_{Na_2SO_4}=0,09(mol)\\ \Rightarrow m_{\downarrow}=m_{BaSO_4}=0,09.233=20,97(g)$