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)?

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Please see # 6  a & b in the upload below.  Thank you

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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)?
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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