Question

The combustion of liquid ethanol (c2H5OH) produces Carbon Dioxide and water. After 4.62 mL of ethanol ( density= 0.789 g/mL) was allowed to burn in the presence of 15.55 g of Oxygen gas, 3.72 mL of water (density=1.00 g/mL) was collected. Determine the limiting reactant, theoretical yield of H20, and percent yield for the reaction. (hint: write a balanced equation for the combustion for the combustion of ethanol.)

Answer 1

THE CHEMICAL EQUESTIONIS

C₂H₅OH (l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)

Next, we need to convert the volumes of ethanol and water to grams using the density formula:

Density = (Mass) / (Volume)
Given that
Density C2H5OH = 0.789 g/mL
Mass C2H5OH = ?
Volume C₂H₅OH= 4.64 mL

Mass C₂H₅OH = (Density C₂H₅OH)(Volume C₂H₅OH)
Mass C₂H₅OH = (0.789 g/mL)(4.64 mL)
Mass C₂H₅OH = 3.66 g C2H5OH
And also
Density of H₂O = 1.00 g/mL
Mass of H₂O = ?
Volume of H₂O = 3.72 mL

Mass H₂O = (Density H₂O)(Volume H₂O)
Mass H₂O = (1.00 g/mL)(3.72 mL)
Mass H₂O = 3.72 g H₂O

Next, we need to use dimensional analysis and equation coefficients to convert 3.66 g of ethanol to moles of ethanol, to moles of water, to grams of water in order to find the theoretical yield.
we know that 1 mol C₂H₅OH = 46.069 g C₂H₅OH
1 mol C₂H₅OH = 3 mol H2O
1 mol H2O = 18.015 g H2O

[(3.66 g of C₂H₅OH)/1]*[(1 mol of C₂H₅OH)/(46.069 g of C₂H₅OH)]*[(3 mol of H₂O)/(1 mol of C₂H₅OH)]*[(18.015 g of H2O)/(1 mol of H2O)] = 4.29 g H₂O

The percent yield of H₂O by using the following equation:

% yield = [(actual yield)(100%)] / (theoretical yield)

% yield = ?
Actual yield = 3.72 g of H₂O (given value by converting ml to moles)
Theoretical yield = 4.29 g of H₂O

% yield = [(3.72 g)(100%)] / (4.29 g)
% yield = 86.7%

Answer:
1. The limiting reagent is ethanol from the balanced equation.
1. The theoretical yield of H₂O for the reaction was 4.29 g.

2. The percent yield for of H₂O for the reaction was 86.7%.