Question

Question 1 2 pts 5.8 g of 1,4-di-t-butyl-2,5-dimethoxynezene (250.373/mol) were synthesized by reacting 10.5 mL oft- butyl alcohol (MW 74.12 3/mol. D 0.795/mL), 25 mL of concentrated sulfuric acid (MW 98.08 5/mol, D 1.84 5/ml), and 5.9 g of 1,4-dimethoxybenzene (MW 138.17 3/mol) together. Calculate the percent yield of this reaction. t-butanol 1,4-di-tert-butyl-2,5-dimethoxybenzene 1,4-dimethoxybenzene Write your answer without a percent sign and to 2 decimal places. MW=molecular weight, D-density of liquid.

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.

Answer 1

percent yield = 54.25 %

Explanation

The balanced chemical equation is : C₈H₁₀O₂ + 2 C₄H₁₀O $\rightarrow$ 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 %