1 0.4212g sample of pure organic acid and required 25.00ml of 0.120N sodium hydroxide(NaOH) for equivalence. Calcaulate the equivalence weight of the acid??
2. In the acid hydrolysis reaction between sucrose( C12H22O11) and water(H2O) which gives glucose(C6H12O6) and fructose( C6H12O6) how many grams of glucose are formed by complete reaction of 32g of sucrose?
3. How many grams ofpotassium dichromate( K2Cr2O7) are repuired to perpare a 250ml solution whose concentration is 2.16N?
THX everybody
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chemical problem(equivalent...
2008-12-21 7:46 pm
回答 (1)
2008-12-21 9:01 pm
✔ 最佳答案
1.No. of eq. of NaOH = NV = 0.12 x (25/1000) = 0.003
No. of eq. of the acid = 0.003
Equivalent weight of the acid = mass/eq. = 0.4212/0.03 = 140.4 g
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2.
C12H22O11 + H2O → C6H12O6(glucose) + C6H12O6(fructose)
Equivalent weight of C12H22O11(sucrose) = 12x12 + 1x22 + 16x11 = 342 g
Equivalent weight of C6H12O6(glucose) = 12x6 + 1x12 + 16x6 = 180 g
No. of eq. of C12H22O11(sucrose) = mass/(eq. mass) = 32/342
No. of eq. of C6H12O6(glucose) = 32/342
Mass of C6H12O6(glucose) = (eq. weight) x (no. of eq.) = 180 x (32/342) = 16.84 g
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3.
If K2Cr2O7 is used as an oxidizing agent: Cr2O72- + 14H2O + 6e- → 2Cr3+ + 7H2O
Equivalent weight of K2Cr2O7 = (39x2 + 52x2 + 16x7)/3 = 98 g
No. of equivalent of K2Cr2O7 = NV = 2.16 x (250/1000) = 0.54
Mass of K2Cr2O7 = (eq. weight) x (no. of eq.) = 98 x 0.54 = 52.92 g
If K2Cr2O7 is used as a precipitating agent:
Equivalent weight of K2Cr2O7 = (39x2 + 52x2 + 16x7)/2 = 147 g
No. of equivalent of K2Cr2O7 = NV = 2.16 x (250/1000) = 0.54
Mass of K2Cr2O7 = (eq. weight) x (no. of eq.) = 147 x 0.54 = 79.38 g
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2008-12-21 14:43:51 補充：
Amendment of Q.3:
Equivalent weight of K2Cr2O7 = (39x2 + 52x2 + 16x7)/6 = 49 g
No. of equivalent of K2Cr2O7 = NV = 2.16 x (250/1000) = 0.54
Mass of K2Cr2O7 = (eq. weight) x (no. of eq.) = 49 x 0.54 = 26.46 g
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