Answer to Question #180952 in Chemistry for sher

Question

Determine the percentage yield of Carbon, when 0.5 mole of Sucrose is decomposed and 70.0 g of C was collected during the experiment. Equation: C12H22O11 = 12 C + 11H2O

Answer 1

Solution:

Balanced chemical equation:

C₁₂H₂₂O₁₁ → 12C + 11H₂O

According to the equation above: n(C₁₂H₂₂O₁₁) = n(C)/12

n(C) = 12 × n(C₁₂H₂₂O₁₁) = 12 × 0.5 mol = 6 mol

Moles of C = 6 mol

Moles of C = Mass of C / Molar mass of C

Mass of C = Moles of C × Molar mass of C

The molar mass of C is 12 g mol^-1.

Hence,

Mass of C = 6 mol × 12 g mol^-1 = 72 g

Mass of C = 72.0 grams - Theoretical Yield

Mass of C = 70.0 grams - Actual Yield

Percentage Yield = (Actual Yield / Theoretical Yield) × 100%

Percentage Yield = (70.0 g / 72.0 g) × 100% = 97.2%

Percentage Yield = 97.2%

Answer: The percentage yield of carbon (C) is 97.2%.