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Câu hỏi: Tính khối lượng muối dùng để sản xuất xà phòng thu được khi cho 100 kg một loại mỡ chứa 50% tristearin; 30% triolein và 20% tripanmitin tác dụng với natri hiđroxit vừa đủ (giả thiết hiệu suất phản ứng đạt 100%).
Phương pháp giải
100 kg mỡ chứa 50% tristearin, 30% triolein và 20% panmitin
$ \Rightarrow {m_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}};{m_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}};{m_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}}$
$ \Rightarrow {n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}};{n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}};{n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}}$
Từ phương trình ta có
${n_{{C_{17}}{H_{35}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} $
${n_{{C_{17}}{H_{33}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} $
${n_{{C_{15}}{H_{31}}COON{\text{a}}}} = 3{n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} $
$ \Rightarrow {m_{muoi}} = {m_{{C_{17}}{H_{35}}COON{\text{a}}}} + {m_{{C_{17}}{H_{33}}COON{\text{a}}}} + {m_{{C_{15}}{H_{31}}COON{\text{a}}}} $
Lời giải chi tiết
Phản ứng của các chất với dung dịch NaOH :
(C17H35COO)3C3H5 + 3NaOH → 3C17H35COONa + C3H5(OH)3
(C17H33COO)3C3H5 + 3NaOH → 3C17H33COONa + C3H5(OH)3
(C15H31COO)3C3H5 + 3NaOH → 3C15H31COONa + C3H5(OH)3
Ta có:
${m_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{100.1000.50}}{{100}} = 50000 gam$
$ \Rightarrow {n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{50000}}{{890}} = 56,18 mol$
${m_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{100.1000.30}}{{100}} = 30000 gam$
$ \Rightarrow {n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{30000}}{{884}} = 33,94 mol$
${m_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{100.1000.20}}{{100}} = 20000 mol$
$ \Rightarrow {n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{20000}}{{806}} = 24,81 mol$
Từ phương trình, ta có:
${n_{{C_{17}}{H_{35}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} = 168,54 mol$
${n_{{C_{17}}{H_{33}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} = 101,82 mol$
${n_{{C_{15}}{H_{31}}COON{\text{a}}}} = 3{n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} = 74,43 mol$
$ \Rightarrow {m_{muoi}} = {m_{{C_{17}}{H_{35}}COON{\text{a}}}} + {m_{{C_{17}}{H_{33}}COON{\text{a}}}} + {m_{{C_{15}}{H_{31}}COON{\text{a}}}} = 168,54.306 + 101,82.304 + 74,43.278$
$ = 103218,06 gam = 103,2 kg$
100 kg mỡ chứa 50% tristearin, 30% triolein và 20% panmitin
$ \Rightarrow {m_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}};{m_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}};{m_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}}$
$ \Rightarrow {n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}};{n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}};{n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}}$
Từ phương trình ta có
${n_{{C_{17}}{H_{35}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} $
${n_{{C_{17}}{H_{33}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} $
${n_{{C_{15}}{H_{31}}COON{\text{a}}}} = 3{n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} $
$ \Rightarrow {m_{muoi}} = {m_{{C_{17}}{H_{35}}COON{\text{a}}}} + {m_{{C_{17}}{H_{33}}COON{\text{a}}}} + {m_{{C_{15}}{H_{31}}COON{\text{a}}}} $
Lời giải chi tiết
Phản ứng của các chất với dung dịch NaOH :
(C17H35COO)3C3H5 + 3NaOH → 3C17H35COONa + C3H5(OH)3
(C17H33COO)3C3H5 + 3NaOH → 3C17H33COONa + C3H5(OH)3
(C15H31COO)3C3H5 + 3NaOH → 3C15H31COONa + C3H5(OH)3
Ta có:
${m_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{100.1000.50}}{{100}} = 50000 gam$
$ \Rightarrow {n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{50000}}{{890}} = 56,18 mol$
${m_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{100.1000.30}}{{100}} = 30000 gam$
$ \Rightarrow {n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{30000}}{{884}} = 33,94 mol$
${m_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{100.1000.20}}{{100}} = 20000 mol$
$ \Rightarrow {n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} = \dfrac{{20000}}{{806}} = 24,81 mol$
Từ phương trình, ta có:
${n_{{C_{17}}{H_{35}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{35}}COO\right)}_3}{C_3}{H_5}}} = 168,54 mol$
${n_{{C_{17}}{H_{33}}COON{\text{a}}}} = 3.{n_{{{\left({C_{17}}{H_{33}}COO\right)}_3}{C_3}{H_5}}} = 101,82 mol$
${n_{{C_{15}}{H_{31}}COON{\text{a}}}} = 3{n_{{{\left({C_{15}}{H_{31}}COO\right)}_3}{C_3}{H_5}}} = 74,43 mol$
$ \Rightarrow {m_{muoi}} = {m_{{C_{17}}{H_{35}}COON{\text{a}}}} + {m_{{C_{17}}{H_{33}}COON{\text{a}}}} + {m_{{C_{15}}{H_{31}}COON{\text{a}}}} = 168,54.306 + 101,82.304 + 74,43.278$
$ = 103218,06 gam = 103,2 kg$