PTHH: (CH316COO)3C3H5 + 3NaOH ---> 3CH316COONa + C3H5(OH)3
Ta có: \(n_{CH_{316}COONa}=\dfrac{45,9}{395}\approx0,12\left(mol\right)\)
Ta lại có: \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{0,15}=1M\)
=> nNaOH = 0,15(mol)
Ta thấy: \(\dfrac{0,12}{3}< \dfrac{0,15}{3}\)
=> NaOH dư.
Theo PT: \(n_{\left(CH_{316}COO\right)_3C_3H_5}=\dfrac{1}{3}.n_{CH_{316}COONa}=\dfrac{1}{3}.0,12=0,04\left(mol\right)\)
=> \(m_{\left(CH_{316}COO\right)_3C_3H_5}=0,04.1157=46,28\left(g\right)\)
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