a, Ta có: \(n_{NaOH}=\dfrac{30.20\%}{40}=0,15\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
__________0,15______0,15 (mol)
\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)
b, \(n_{Na_2CO_3}=0,2.0,5=0,1\left(mol\right)\)
PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,1}{1}\), ta được Na2CO3 dư.
Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)