Mr.Q 發問時間: 科學化學 · 9 年前

普化 化學計量 問題

Consider the fermentation reaction of glucose:

C6H12O6 → 2C2H5OH + 2CO2

A 1.00 mole sample of C6H12O6 was placed in a vat with 100 g of yeast. If 57.4 grams of C2H5OH was obtained, what was the percent yield of C2H5OH ?

(A) 62.3 %

(B) 31.1 %

(C) 100 %

(D) 57.4 %

(E) none of these

已更新項目:

Answer:(A)

2 個已更新項目:

選項貼錯 = = 抱歉~

更正:

 (A)50 %

 (B)56 %

 (C)100 %

 (D)42 %

 (E)none of these

1 個解答

評分
  • Adam
    Lv 7
    9 年前
    最佳解答

    The original options arecorrect, but the new options are NOT.

    The answer is : (A) 62.3%

    C6H12O6 → 2C2H5OH+ 2CO2

    No. of moles of C6H12O6 = 1 mol

    Molar mass of C2H5OH = 12x2 + 1x6 + 16 = 46 g/mol

    Theoretical (maximum) yield of C2H5OH = 1 x 2 x 46 = 92 g

    Percentage yield of C2H5OH = (57.4/92) x 100% = 62.39% ≈62.3%

    參考資料: Adam
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