Answer to Question #180952 in Chemistry for sher

Question #180952

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


1
Expert's answer
2021-04-14T06:11:59-0400

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%.

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