\(n_{NaOH}=\dfrac{42}{40}=1,05\left(mol\right)\)
PTHH: (RCOO)3C3H5 + 3NaOH ---> 3RCOONa + C3H5(OH)3
1,05---------------------------->0,35
\(m_{C_3H_5\left(OH\right)_3}=0,35.92=32,2\left(g\right)\)
Bảo toàn khối lượng:
\(m_{chất.béo}+m_{NaOH}=m_{xà.phòng}+m_{C_3H_5\left(OH\right)_3}\\ \Rightarrow m_{xà.phòng}=89+42-32,2=98,8\left(g\right)\)