A. 0,18
B. 0,21
C. 0,24
D. 0,27
B
\( \to \left( \begin{array}{l} {\mkern 1mu} 42,{\mkern 1mu} 38{\mkern 1mu} gam{\mkern 1mu} X \to \left\{ \begin{array}{l} Z{\mkern 1mu} (RCOOK) \to \left. \begin{array}{l} 152,63{\mkern 1mu} gam\left\{ \begin{array}{l} {\mkern 1mu} C{O_2}\\ {H_2}O \end{array} \right.\\ {K_2}C{O_3} \end{array} \right\rangle \\ 26,2{\mkern 1mu} gam{\mkern 1mu} Y\left\{ \begin{array}{l} {C_3}{H_5}{(OH)_3}\\ {H_2}O \end{array} \right.{\mkern 1mu} \end{array} \right.\\ 0,15{\mkern 1mu} mol{\mkern 1mu} X \to \end{array} \right.\)
nglyxeron = nx= nglyxeron = x
→ nKOH =3nx → nKOH = 3x → mKOH =168x → mddKOH =168x/28.100= 600x → mH2O = 600x-168x=432x
Theo đề: mH2O + mglyxeron = 26,2 → x=0,05 mol
432 92
Ta có: nKOH = 0,15 mol → 42,38 + 0,15.56 = mRCOOK + 0,05.92 → mRCOOK = 46,18 gam
Phương trình đốt cháy muối: \( \begin{align} & 2{{C}_{x}}{{H}_{y}}{{O}_{2}}K+\left( \left. 2x+\frac{y}{2}-1 \right) \right.{{O}_{2}}\xrightarrow{{}}(2x-1)C{{O}_{2}}\,\,+\,\,\,y{{H}_{2}}O+{{K}_{2}}C{{O}_{3}} \\ & \,\,\,\,\,\,\,\,0,15\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2x-1).0,075\,\,\,\,\,0,075y \\ \end{align}\)
Ta có: lk trong -C- C- = nBr2 /nx → nBr2= 0,21
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247