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Answer:
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Explanation:
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C₄H₉OH + HBr = C₄H₉Br + H2O
Δmole of alcohol gives 1 mole of bromobutanol
HBr is in excess, so the yield of the product is limited by the alcohol
Wt. of 1 butanol = 18
Molar mass of the butanol = 74.12 g/mole
Moles of the alcohol = 1/74.12 = 0.01349 moles
So, moles of bromobutane = 0.01349 moles
Molar mass of C₄H₉Br = 137.018 g/moles
So, theoretical mass of bromobutane is = 0.01349 × 137.0.18
= 1.85 g
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