5.8 g of 1,4-di-t-butyl-2,5-dimethoxynezene (250.37 g/mol) were synthesized by reacting 10.5 mL of t-butyl alcohol (MW 74.12 g/mol, D 0.79 g/mL), 25 mL of concentrated sulfuric acid (MW 98.08 g/mol, D 1.84 g/mL), and 5.9 g of 1,4-dimethoxybenzene (MW 138.17 g/mol) together. Calculate the percent yield of this reaction.
percent yield = 54.25 %
Explanation
The balanced chemical equation is : C8H10O2 + 2 C4H10O C16H26O2 + 2 H2O
Case 1 : t-butyl alcohol C4H10O is the limiting reagent
volume t-butyl alcohol C4H10O = 10.5 mL
mass C4H10O = (volume C4H10O) * (density C4H10O)
mass C4H10O = (10.5 mL) * (0.79 g/mL)
mass C4H10O = 8.295 g
moles C4H10O = (mass C4H10O) / (molar mass C4H10O)
moles C4H10O = (8.295 g) / (74.12 g/mol)
moles C4H10O = 0.112 mol
moles C16H26O2 produced = (1/2) * (moles C4H10O)
moles C16H26O2 produced = (0.5) * (0.112 mol)
moles C16H26O2 produced = 0.056 mol
mass C16H26O2 produced = (moles C16H26O2 produced) * (molar mass C16H26O2)
mass C16H26O2 produced = (0.056 mol) * (250.37 g/mol)
mass C16H26O2 produced = 14.01 g
Case 2 : 1,4-dimethoxybenzene is the limiting reagent
mass 1,4-dimethoxybenzene C8H10O2 = 5.9 g
moles C8H10O2 = (mass C8H10O2) / (molar mass C8H10O2)
moles C8H10O2 = (5.9 g) / (138.17 g/mol)
moles C8H10O2 = 0.0427 mol
moles C16H26O2 produced = moles C8H10O2
moles C16H26O2 produced = 0.0427 mol
mass C16H26O2 produced = (moles C16H26O2 produced) * (molar mass C16H26O2)
mass C16H26O2 produced = (0.0427 mol) * (250.37 g/mol)
mass C16H26O2 produced = 10.69 g
Since less mass of C16H26O2 is produced in case 2, therefore, C8H10O2 is the limiting reagent.
theoretical yield C16H26O2 = 10.69 g
percent yield = (actual yield C16H26O2 / theoretical yield C16H26O2) * 100
percent yield = (5.8 g / 10.69 g) * 100
percent yield = 54.25 %
5.8 g of 1,4-di-t-butyl-2,5-dimethoxynezene (250.37 g/mol) were synthesized by reacting 10.5 mL of t-butyl alcohol (MW...
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