7)
Phản ứng xảy ra:
\({({C_{17}}{H_{33}}COO)_3}{C_3}{H_5} + 3{H_2}\xrightarrow{{Ni,{t^o}}}{({C_{17}}{H_{35}}COO)_3}{C_3}{H_5}\)
Ta có:
\({n_{{{({C_{17}}{H_{35}}COO)}_3}{C_3}{H_5}}} = \frac{5}{{(12.17 + 35 + 44).3 + 12.3 + 5}} = \frac{5}{{890}}\)
Ta có:
\({n_{{{({C_{17}}{H_{33}}COO)}_3}{C_3}{H_5}}} = {n_{{{({C_{17}}{H_{35}}COO)}_3}{C_3}{H_5}}} = \frac{5}{{890}}\)
\( \to {m_{{{({C_{17}}{H_{33}}COO)}_3}{C_3}{H_5}}} = \frac{5}{{890}}.((12.17 + 33 + 44).3 + 12.3 + 5) = 4,966292{\text{ tấn = 4966}}{\text{,292 kg}}\)
Chọn \(A\).
8)
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:
\({m_{tristearin}} = 4,45.(100\% - 20\% ) = 3,56{\text{ gam}}\)
\( \to {n_{tristearin}} = \frac{{3,56}}{{890}} = {n_{{C_3}{H_5}{{(OH)}_3}}}_{lt}\)
\( \to {m_{{C_3}{H_5}{{(OH)}_3}{\text{ (lt)}}}} = \frac{{3,56}}{{890}}.92 = 0,368{\text{ gam}}\)
\( \to {m_{{C_3}{H_5}{{(OH)}_3}}} = 0,368.85\% = 0,3128{\text{ gam}}\)
Chọn \(A\)
9)
Sơ đồ phản ứng:
\(X + NaOH\xrightarrow{{}}muối + {C_3}{H_5}{(OH)_3}\)
\({n_{{C_3}{H_5}{{(OH)}_3}}} = \frac{{9,2}}{{92}} = 0,1{\text{ mol}} \to {n_{muối}} = 3{n_{{C_3}{H_5}{{(OH)}_3}}} = 0,3{\text{ mol}}\)
Gọi \(B\) có dạng \(C_nH_{2n+1}COOH\) suy ra muối là \(C_nH_{2n+1}COONa\)
\( \to 0,3.(12n + 2n + 1 + 44 + 23) = 83,4 \to n = 15\)
Vậy \(B\) là \(C_{15}H_{31}COOH\)
Suy ra \(B\) là axit panmitic.
