Answer – Given, [H2SO4] = 8.7 M , volume = 25 mL, volume of 1-butanol = 7 mL
Density of 1-butanol = 0.81 g/mL , mass of NaBr = 10.4 g
First we need to write the reaction
CH3CH2CH2CH2OH + NaBr + H2SO4 ------> CH3CH2CH2CH2Br + NaOH + H2SO4
We need to calculate the moles of each
Moles of H2SO4 = 8.7 M * 0.025 L = 0.218 moles
Moles of NaBr = 10.4 g / 102.894 g.mol-1
= 0.101 moles
Mass of 1-butanol = 7 mL * 0.81 g/mL = 5.67 g
Moles of 1-butanol = 5.67 g / 74.12 g.mol-1 = 0.0765 moles
We know all has mole ratio is 1:1 with each other
So lowest moles of 1-bromobutane formed from the 1-butanol, since 1-butanol is limiting reactant.
So theoretical yield of 1-bromobutane = 0.0765 moles * 137.018 g/mol
= 10.5 g
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