A student reascts 4.00 g of aniline (MW=93.14 g/mole) with 5..40 mL of acetic anhydride (MW=102.07 g/mole, p=1.082 g/mL) which produces 4.23 g of acetanilide (MW= 135.18 g/mole)
1) HOW MANY MOLES OF ANILINE ARE USED IN THE REACTION?
2) WHAT MASS OF ACCETIC ANHYDRIDE IS USED IN THE REACTION?
3)HOW MANY MOLES OF ACCETIC ANHYDRIDE ARE USED IN THE REACTION ??
Please HELP
Chemistry help?
2017-09-04 12:20 pm
回答 (2)
2017-09-04 1:13 pm
1)
The balanced equation of the reaction :
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
(aniline) + (acetic anhydride) → (acetanilide) + (acetic acid)
Mole ratio aniline : acetanilide = 1 : 1
No. of moles of acetanilide produced = (4.23 g) / (135.18 g/mol) = 0.0313 mol
No. of moles of aniline reacted = No. of moles of acetanilide produced = 0.0313 mol
2)
According to the balanced equation, mole ratio acetic anhydride : acetanilide = 1 : 1
No. of moles of acetanilide produced = (4.23 g) / (135.18 g/mol) = 0.0313 mol
No. of moles of acetic anhydride reacted = No. of moles of acetanilide produced = 0.0313 mol
Mass of acetic anhydride reacted = (0.0313 mol) × (102.07 g/mol) = 3.19 g
3)
Refer to Part 2).
No. of moles of acetic anhydride reacted = 0.0313 mol
The balanced equation of the reaction :
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
(aniline) + (acetic anhydride) → (acetanilide) + (acetic acid)
Mole ratio aniline : acetanilide = 1 : 1
No. of moles of acetanilide produced = (4.23 g) / (135.18 g/mol) = 0.0313 mol
No. of moles of aniline reacted = No. of moles of acetanilide produced = 0.0313 mol
2)
According to the balanced equation, mole ratio acetic anhydride : acetanilide = 1 : 1
No. of moles of acetanilide produced = (4.23 g) / (135.18 g/mol) = 0.0313 mol
No. of moles of acetic anhydride reacted = No. of moles of acetanilide produced = 0.0313 mol
Mass of acetic anhydride reacted = (0.0313 mol) × (102.07 g/mol) = 3.19 g
3)
Refer to Part 2).
No. of moles of acetic anhydride reacted = 0.0313 mol
2017-09-04 12:39 pm
It's Acetic anhydride.
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