a, \(m_{CH_3COOH}=20.3,75\%=0,75\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{0,75}{60}=0,0125\left(mol\right)\)
PT: \(2CH_3COOH+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2H_2O\)
Theo PT: \(n_{Ca\left(OH\right)_2}=\dfrac{1}{2}n_{CH_3COOH}=0,00625\left(mol\right)\)
\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{0,00625}{0,2}=0,03125\left(l\right)=31,25\left(ml\right)\)
b, \(n_{\left(CH_3COO\right)_2Ca}=\dfrac{1}{2}n_{CH_3COOH}=0,00625\left(mol\right)\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Ca}=0,00625.158=0,9875\left(g\right)\)