a) PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\) (1)
Ta có: \(n_{NaOH}=0,05\cdot2=0,1\left(mol\right)\) \(\Rightarrow n_{H_2SO_4}=0,05mol\)
\(\Rightarrow m_{H_2SO_4}=0,05\cdot98=4,9\left(g\right)\) \(\Rightarrow m_{ddH_2SO_4}=\frac{4,9}{20\%}=24,5\left(g\right)\)
b) PTHH: \(BaCl_2+Na_2SO_4\rightarrow2NaCl+BaSO_4\downarrow\) (2)
Theo PTHH (1): \(n_{Na_2SO_4}=n_{H_2SO_4}=0,05mol\)
Theo PTHH (2): \(n_{BaCl_2}=\frac{20,8}{208}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\frac{0,05}{1}< \frac{0,1}{1}\) \(\Rightarrow\) Na2SO4 phản ứng hết, BaCl2 còn dư
\(\Rightarrow n_{BaSO_4}=0,05mol\) \(\Rightarrow m_{BaSO_4}=0,05\cdot233=11,65\left(g\right)\)