6) a) 74.0 g of butan-1-ol (C4H9OH) reacting with excess hydrogen bromide will produce how many grams of C4H9Br? C4H9QH (1) + HBr (g) → C4H9Br (1) + H₂O (1) 74. од Сту на он 20 4 1- (gH ₂ OH | |-Luths 137.03 9 (4 B 1 74.149 1 - Cuttyor |- Cy Hg Br. 10, 140.22 74.14 48.04+ 9,09 + 16 +1.01. b) If 15.0 g of C4H9Br was actually obtained, what is the percent yield (what percent of what the reaction should have produced was actually produced)?
6) a) 74.0 g of butan-1-ol (C4H9OH) reacting with excess hydrogen bromide will produce how many grams of C4H9Br? C4H9QH (1) + HBr (g) → C4H9Br (1) + H₂O (1) 74. од Сту на он 20 4 1- (gH ₂ OH | |-Luths 137.03 9 (4 B 1 74.149 1 - Cuttyor |- Cy Hg Br. 10, 140.22 74.14 48.04+ 9,09 + 16 +1.01. b) If 15.0 g of C4H9Br was actually obtained, what is the percent yield (what percent of what the reaction should have produced was actually produced)?
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter4: Chemical Reactions
Section: Chapter Questions
Problem 4.55P: 4-55 For the reaction: (a) How many moles of N2 are required to react completely with 1 mole of O2?...
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![3
6) a) 74.0 g of butan-1-ol (C4H₂OH) reacting with excess hydrogen bromide will produce how many grams
C4H9QH (1)
+
HBr (g) →
C4H9Br (1)
of C4H9Br?
H₂O (1)
4 CH
1-C4H₂ OH | |-Luths 137.03
14. од Сту на он
74.14g |· 1-Cuthar | 1-C₂ H₂ Br
Cy Hq
+
ż
10, 140.22
74.14
48.04 + 9,09 + 16 +1.01
b) If 15.0 g of C4H9Br was actually obtained, what is the percent yield (what percent of what the reaction
should have produced was actually produced)?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bf2a401-dbe7-4037-940c-9cf04f156659%2Ffbb62e03-a991-4a1d-b27b-6d8b012a2705%2Fble59jg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3
6) a) 74.0 g of butan-1-ol (C4H₂OH) reacting with excess hydrogen bromide will produce how many grams
C4H9QH (1)
+
HBr (g) →
C4H9Br (1)
of C4H9Br?
H₂O (1)
4 CH
1-C4H₂ OH | |-Luths 137.03
14. од Сту на он
74.14g |· 1-Cuthar | 1-C₂ H₂ Br
Cy Hq
+
ż
10, 140.22
74.14
48.04 + 9,09 + 16 +1.01
b) If 15.0 g of C4H9Br was actually obtained, what is the percent yield (what percent of what the reaction
should have produced was actually produced)?
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