The combustion of liquid ethanol (c2h5oh) produces carbon dioxide and water. after 4.61 ml of ethanol (density=0.789g/ml) was allowed to burn in the presence of 15.70 g of oxygen gas, 3.72 ml of water (density=1.00g/ml) was collected. part a determine the percent yield of h2o for the reaction.

1) Chemical equation

C2H5OH + 3 O2 ---> 2 CO2 + 3 H2O

2) Molar ratios

1 mol C2H5OH : 3 mol O2 : 2 mol CO2 : 3 mol H2O

3) Amount of ethanol burned

D = M / V => M = D * V = 0.789 g/ml * 4.61 ml = 3.637 g

Number of moles = mass in grams / molar mass

molar mass of C2H5OH = 2 *12 g/mol + 6*1.0 g/mol + 16.0 g/mol = 46.0 g/mol

=> number of moles of C2H5OH = 3.637 g / 46.0 g/mol = 0.079 mol

4) Amount of oxygen, O2

number of moles of O2 = mass in grams / molar mass

number of moles O2 = 15.70g / 32.0 g/mol = 0.49 mol

5) Limiting reactant

Theoretical ratio: 1 mol C2H5OH / 3 mol O2

Actual ratio: 0.079 mol / 0.49 mol = 0.16

=> there is more oxygen than needed to burn all the ethanol => ethanol burns completely and it is the limiting reactant.

6) Theoretical yield of H2O

1 mol C2H5OH / 3 mol H2O = 0.079 mol C2H5OH / x

=> x = 0.079 mol C2H5OH * 3 mol H2O / 1 mol C2H5OH

=> x = 0.237 mol H2O

Conversion to grams:

mass = molar mass * number of moles

mass H2O = 18.0 g/mol * 0.237 g/mol = 4.27 g < theoretical yield

7) grams of H2O collected (yield)

M = V * D = 3.27 ml * 1.00 g/ml = 3.27 g < actual yield

8) Percent yield

Percent yield = (actual yield / theoretical yield) * 100 = (3.27 g / 4.27) * 100 = 76.6%

76.6%

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