Câu `3:`

`n_{NaOH}=5,25.1=5,25(mol)`

`=>n_{OO}=n_{\text{gốc COO ester}}+n_{\text{gốc COOH acid}}=n_{NaOH}=5,25(mol)`

`=>n_{O(E)}=5,25.2=10,5(mol)`

`=>m_{O(E)}=10,5.16=168(g)`

`m_C=76,2\% . 1468,5\approx 1119(g)`

`=>m_H=1468,5-168-1119=181,5(g)`

`\%H=(181,5)/(1468,5).100\%\approx 12,36\%`

Câu `4:`

`n_{C_3H_5(OH)_3}=(115)/(92)=1,25(mol)`

`=>n_{COO(\text{ester})}=1,25.3=3,75(mol)`

`=>n_{COO(\text{acid})}=5,25-3,75=1,5(mol)`

`=>n_{\text{acid}}=1,5(mol)`

Câu `5:`

`n_{H_2O(\text{sản phẩm})}=n_{\text{acid}}=1,5(mol)`

Bảo toàn khối lượng :

`1468,5+5,25.40=a+115+1,5.18`

`=>a=1536,5(g)`