`m_{\text{triolein}}=5,525.80\%=4,42(kg)`

`n_{\text{triolein}}=(4,42)/(884)=0,005(kmol)`

PTHH:

`(C_{17}H_{33}COO)_3C_3H_5+3NaOH\to 3C_{17}H_{33}COONa+C_3H_5(OH)_3`

`=>n_{C_{17}H_{33}COONa}=3.0,005=0,015(kmol)` hay `15(mol)`

`=>m_{C_{17}H_{33}COONa}=15.304=4560(g)`

`=>x=(4560)/(60)=76`

$\rm m_{triolein}=5,525.80\%=4,42(kg)$

`->`$\rm n_{triolein}=\dfrac{4,42}{884}=0,005(kmol)$ 

Phương trình hóa học 

$\rm (C_{17}H_{33}COO)_3C_3H_5 +3NaOH \xrightarrow{H_2O,to} 3C_{17}H_{33}COONa +C_3H_5(OH)_3$

Theo phương trình hóa học 

$\rm n_{sodium oleate}=3n_{(C_{17}H_{33}COO)_3C_3H_5}=0,015(kmol)$ 

`->`$\rm m_{sodium oleate}=0,015.304=4,65(kg)=4560(g)$

Số bánh xà phòng 

$\rm x=\dfrac{4560}{60}=76(bánh)$