Đáp án đúng: A
Lời giải:${{\text{n}}_{{\text{C}{{\text{O}}_{\text{2}}}}}}\,\text{=}\,\text{4,275}\,\text{mol,}\,{{\text{n}}_{{{{\text{H}}_{\text{2}}}\text{O}}}}\,\text{=}\,\text{3,9}\,\text{mol}$Bảo toàn khối lượng trong X ⇒${{\text{m}}_{\text{O}}}\,\text{=}\,{{\text{m}}_{\text{X}}}\,\text{-}\,{{\text{m}}_{\text{C}}}\,\text{-}\,{{\text{m}}_{\text{H}}}$= 66,3 – 4,275.12 – 3,9.2 =7,2⇒${{\text{n}}_{\text{O}}}$= 0,45 mol ⇒${{\text{n}}_{\text{X}}}\,\text{=}\,\frac{{{{\text{n}}_{\text{O}}}}}{\text{6}}$= 0,075 mol ⇒ X có công thức dạng${{\text{C}}_{{\text{51}}}}{{\text{H}}_{{\text{99}}}}{{\text{(COO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}$⇒${{\text{n}}_{{{{\text{H}}_{\text{2}}}}}}$= 0,15 mol ⇒${{\text{n}}_{\text{X}}}\,\text{=}\,{{\text{n}}_{\text{Y}}}\,\text{=}\,\frac{{{{\text{n}}_{\text{Y}}}}}{\text{3}}$= 0,05 mol. Y có công thức${{\text{C}}_{{\text{51}}}}{{\text{H}}_{{\text{105}}}}{{\text{(COO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}$
${{\text{C}}_{{\text{51}}}}{{\text{H}}_{{105}}}{{\text{(COO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}$
+
3KOH
→
${{\text{C}}_{{\text{51}}}}{{\text{H}}_{{105}}}{{\text{(COOK)}}_{\text{3}}}$
+
${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}{{\text{(OH)}}_{\text{3}}}$
Ban đầu
0,05 mol
0,5 mol
Phản ứng
0,05 mol
→
0,15 mol
→
0,05 mol
→
0,05 mol
Sau phản ứng
0,35 mol
0,05 mol
0,05 mol
Khối lượng rắn là b = 0,35.56 + 0,05.966 = 67,9 gam. Đáp án A.
