1)
Phản ứng xảy ra:
\({({C_{17}}{H_{35}}COO)_3}{C_3}{H_5} + 3NaOH\xrightarrow{{}}3{C_{17}}{H_{35}}COONa + {C_3}{H_5}{(OH)_3}\)
Ta có:
\({n_{{{({C_{17}}{H_{35}}COO)}_3}{C_3}{H_5}}} = \frac{{26,7}}{{283.3 + 41}} = 0,03{\text{ kmol = }}{{\text{n}}_{{C_3}{H_5}{{(OH)}_3}}} \to {m_{{C_3}{H_5}{{(OH)}_3}}} = 0,03.92 = 2,76{\text{ kg}}\)
Chọn đáp án B.
2)
Chất béo có dạng \({(RCOO)_3}{C_3}{H_5}\)
\({(RCOO)_3}{C_3}{H_5} + 3NaOH\xrightarrow{{}}RCOONa + {C_3}{H_5}{(OH)_3}\)
BTKL:
\({m_{{{(RCOO)}_3}{C_3}{H_5}}} + {m_{NaOH}} = {m_{muối}} + {m_{{C_3}{H_5}{{(OH)}_3}}} \to {m_{muối}} = 17,16 + 2,4 - 1,84 = 17,72{\text{ kg}}\)
Chọn A
3)
\({({C_{17}}{H_{35}}COO)_3}{C_3}{H_5} + 3NaOH\xrightarrow{{}}3{C_{17}}{H_{35}}COONa + {C_3}{H_5}{(OH)_3}\)
Ta có:
\({n_{{C_{17}}{H_{35}}COONa}} = \frac{1}{{306}} \to {n_{{{({C_{17}}{H_{35}}COO)}_3}{C_3}{H_5}{\text{ lý thuyết}}}} = \frac{{\frac{1}{{306}}}}{3} = \frac{1}{{918}} \to {n_{{{({C_{17}}{H_{35}}COO)}_3}{C_3}{H_5}{\text{ thực tế}}}}{\text{ = }}\frac{1}{{918.80\% }} = \frac{5}{{3672}}\)
\( \to {n_{{{({C_{17}}{H_{35}}COO)}_3}{C_3}{H_5}{\text{ thực tế}}}} = \frac{5}{{3672}}.890 = 1,212{\text{ tấn}}\)
Chọn D.