a) PTHH: \(2CH_3COOH+CaO\rightarrow\left(CH_3COO\right)_2Ca+H_2O\)
b) Ta có: \(n_{CH_3COOH}=\frac{15}{60}=0,25\left(mol\right)\)
\(\Rightarrow n_{CaO}=0,125mol\) \(\Rightarrow m_{CaO}=0,125\cdot56=7\left(g\right)\)
c) Theo PTHH: \(n_{CaO}=n_{\left(CH_3COO\right)_2Ca}=0,125mol\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Ca}=0,125\cdot158=19,75\left(g\right)\)
\(\Rightarrow mThực_{\left(CH_3COO\right)_2Ca}=19,75\cdot85\%=16,7875\left(g\right)\)