Câu 47.5*
\(n_{C_3H_5\left(OH\right)_3}=\dfrac{1,84}{92}=0,02mol\)
\(\left(RCOO\right)C_3H_5+3NaOH\rightarrow C_3H_5\left(OH\right)_3+3RCOONa\)
1 3 1 3 (theo tỉ lệ)
0,06 0,02
\(m_{NaOH}=0,06\cdot40=2,4g\)
Bảo toàn khối lượng:
\(m_{chấtbéo}+m_{NaOH}=m_{muối}+m_{glixerol}\)
\(\Rightarrow m+2,4=17,72+1,84\)
\(\Rightarrow m=17,16g\)