emistry com/imyct/itemiewassignmentProblemiD-981233 CH 8 HW Exercise 8.92 18 of

Question

emistry com/imyct/itemiewassignmentProblemiD-981233 CH 8 HW Exercise 8.92 18 of 20 Constants Eecod Part A The combustion of liquid ethanol (CoHs OH) produces carbon doxide and water. After 3.8 mL of ethanol (density 0.789 g/mL) was allowed to bum in the presence of 12.5 g of oxygen gas, 3 10 mL of water (density 1.00 g/mL) was collected Determine the limiting reactand(Hint.Wnite a balanced equation for the combustion of ethanol) Express your answer as a chemical formula Submit Part B Determine the theoresical yield of H O Express your answer using two significant figures 20

Explanation / Answer

C2H5OH (l) + 3 O2 (g) ---------> 2 CO2 (g) + 3 H2O (g)

Mass of ethanol = density * volume = 0.789 * 3.8 = 3.00 g.

Molar mass of ethanol = 46.0 g/mol

Moles of ethanol = mass / molar mass = 3.00 / 46.0 = 0.0652 mol

Mass of O2 gas = 12.5 g.

Molar mass of O2 gas = 32.0 g/mol

Moles of O2 gas = mass / molar mass = 12.5 / 32.0 = 0.391 mol

(a)

From the balanced equation,

1 mol of ethanol needs 3 mol of O2

Then, 0.0652 mol of ethanol needs 0.0652 * 3 = 0.196 mol of O2 ( < 0.396 )
Hence ETHANOL is limiting reagent.

(b)

From the balanced equation,

1 mol of ethanol produces 3 mol of H2O
then, 0.0652 mol of ethanol 3 * 0.0652 = 0.196 mol of H2O

Then, Mass of H2O produced = theoretical yield of H2O = 0.196 * 18.0 = 3.52 g.

(c)

Actual yield of H2O = density * volume = 1.00 * 3.10 = 3.10 g.

% Yield = Actual yield * 100 / Theoretical yield

% yield = 31.0 * 100 / 3.52 = 88.0 %

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