Question

The combustion of liquid ethanol (C₂H₅OH) produces carbon dioxide and water. After 4.61 mL of ethanol (density=0.789gml−1) was allowed to burn in the presence of 15.60 g of oxygen gas, 3.73 mL of water (density=1.00gml−1) was collected.

Part A

Determine the limiting reactant for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.)

Determine the limiting reactant for the reaction. (: Write a balanced equation for the combustion of ethanol.)

ethanol

oxygen

Part B

Determine the theoretical yield of H2O for the reaction.

Express your answer using three significant figures.r

Part C

Determine the percent yield of H2O for the reaction.

Express your answer using three significant figures.

Answer 1

Answer:

The balanced reaction for the combustion of ethanol is

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

Given volume of ethanol=4.61 mL and density=0.789 g/mL

Therefore mass of ethanol=density x volume=0.789 g/mL x 4.61 mL =3.637 g.

Molar mass of ethanol=46.07 g/mol

Moles of ethanol=mass/molar mass=3.637 g/46.07 g/mol=0.07895 mol

Mass of O2(g)=15.60 g

Molar mass of O2=32 g/mol.

Moles of O2=15.60 g/32 g/mol=0.4875 mol.

From the balanced equation, the mole ratio between

Ethanol:O2=1:3

But here, ethanol:O2=0.07895:0.4875=1:6.17.

Therefore O2(g) is in excess.

Part A:

The limiting reactant is Ethanol.

Part B:

The moles of H2O formed=3 x moles of ethanol=3 x 0.07895 mol=0.23685 mol.

Molar mass of H2O=18g/mol

Theoretical yield of H2O formed=moles x molar mass=0.23685 mol x 18 g/mol=4.263 g ~ 4.26 g.

Part C:

Given volume of H2O=3.73 mL and density=1 g/mL

Given experimental yield=3.73 mL x 1 g/mL=3.73 g.

Therefore % yield=(experimental mass/theoretical mass) x 100

% yield=(3.73 g/4.26 g) x 100=87.49 %.

% yield ~ 87.5 %.

Please let me know if you have any doubt. Thanks.