Respuesta :
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
Answer:
Theoretical yield = 1.85 g
Explanation:
Moles of [tex]1-butanol[/tex]:-
Mass = 1.0 g
Molar mass of [tex]1-butanol[/tex] = 74.12 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.0\ g}{74.12\ g/mol}[/tex]
[tex]Moles_{1-butanol}= 0.0135\ mol[/tex]
According to the reaction:-
C₄H₉OH + HBr ⇒ C₄H₉Br + H₂O
1 mole of 1-butanol on reaction forms 1 mole of 1-bromobutane
0.0135 mole of 1-butanol on reaction forms 0.0135 mole of 1-bromobutane
Mole of 1-bromobutane = 0.0135 mole
Molar mass of 1-bromobutane = 137.02 g/mol
Mass = Moles * Molar mass = 0.0135 mole * 137.02 g/mol = 1.85 g
Theoretical yield = 1.85 g
