The combustion of liquid ethanol (C2H5OH) produces carbon dioxide and water. After 4.60mL of ethanol (density=0.789g/ml) was allowed to burn in the presence of 15.45g of oxygen gas, 3.70mL of water (density=1.00g/ml) was collected.

a)Determine the theoretical yield of H2O for the reaction.

b)Determine the percent yield of H2O for the reaction.

a) Well, to determine the theoretical yield of H2O for the reaction, we need to calculate the amount of water that can be produced based on the stoichiometry of the reaction. The balanced equation for the combustion of ethanol is:

C2H5OH + 3O2 -> 2CO2 + 3H2O

According to the equation, for every 1 mole of ethanol burned, we get 3 moles of water. So, let's calculate the moles of ethanol burned first:

moles of ethanol = volume of ethanol (in L) x density of ethanol (g/mL) / molar mass of ethanol

moles of ethanol = 4.60 mL * 0.789 g/mL / 46.07 g/mol

moles of ethanol = 0.07862 mol

Now, since the ratio of ethanol to water is 1:3, the moles of water produced will be 3 times the moles of ethanol burned:

moles of water = moles of ethanol * 3

moles of water = 0.07862 mol * 3

moles of water = 0.2359 mol

Now, to convert moles of water to grams, we can use the molar mass of water:

mass of water = moles of water * molar mass of water

mass of water = 0.2359 mol * 18.015 g/mol

mass of water = 4.249 g

Therefore, the theoretical yield of H2O for the reaction is approximately 4.249 grams.

b) To determine the percent yield of H2O, we need to compare the actual yield (which is given as 3.70 mL) to the theoretical yield (which we calculated as 4.249 g).

To compare volumes, we can use the relationship between volume and density:

actual yield of water = volume of water x density of water

actual yield of water = 3.70 mL * 1.00 g/mL

actual yield of water = 3.70 g

Now, we can calculate the percent yield using the formula:

percent yield = (actual yield / theoretical yield) x 100%

percent yield = (3.70 g / 4.249 g) x 100%

percent yield = 87.02%

Therefore, the percent yield of H2O for the reaction is approximately 87.02%.

To determine the theoretical yield of H2O for the reaction, we need to calculate the amount of moles of ethanol and oxygen reacted and use the stoichiometry of the balanced chemical equation.

The balanced chemical equation for the combustion of ethanol is:

C2H5OH + 3O2 -> 2CO2 + 3H2O

a) Calculating the moles of ethanol:
Given volume of ethanol = 4.60 mL
Density of ethanol = 0.789 g/mL
Mass of ethanol = volume x density = 4.60 mL x 0.789 g/mL = 3.6284 g

Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Moles of ethanol = mass of ethanol / molar mass of ethanol
Moles of ethanol = 3.6284 g / 46.07 g/mol = 0.0788 mol

b) Calculating the moles of oxygen:
Given mass of oxygen = 15.45 g
Molar mass of oxygen (O2) = 32.00 g/mol
Moles of oxygen = mass of oxygen / molar mass of oxygen
Moles of oxygen = 15.45 g / 32.00 g/mol = 0.483 mol

According to the balanced equation, the ratio of moles of water to moles of ethanol is 3:1. So if 0.0788 moles of ethanol react completely, the number of moles of water produced will be 3 x 0.0788 mol = 0.2364 mol.

To calculate the theoretical yield of water:
Molar mass of water (H2O) = 18.02 g/mol
Mass of water = moles of water x molar mass of water
Mass of water = 0.2364 mol x 18.02 g/mol = 4.2576 g

Therefore, the theoretical yield of H2O for the reaction is 4.2576 g.

b) To calculate the percent yield of H2O:
Given volume of water collected = 3.70 mL
Density of water = 1.00 g/mL
Mass of water collected = volume x density = 3.70 mL x 1.00 g/mL = 3.70 g

Percent yield = (actual yield / theoretical yield) x 100%
Percent yield = (3.70 g / 4.2576 g) x 100% = 86.85%

Therefore, the percent yield of H2O for the reaction is approximately 86.85%.

To determine the theoretical yield of H2O for the reaction, we first need to balance the chemical equation for the combustion of ethanol:

C2H5OH + O2 -> CO2 + H2O

The balanced equation shows that for every 1 mole of ethanol (C2H5OH), we get 1 mole of water (H2O) as the product.

a) Determine the theoretical yield of H2O for the reaction:

First, we need to calculate the number of moles of ethanol used:

Mass of ethanol = volume x density = 4.60 mL x 0.789 g/mL = 3.6294 g

Molar mass of ethanol (C2H5OH) = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.07 g/mol

Number of moles of ethanol = mass of ethanol / molar mass of ethanol = 3.6294 g / 46.07 g/mol = 0.0787 mol

Since the stoichiometry of the balanced equation shows that for every 1 mole of ethanol, we obtain 1 mole of water, the theoretical yield of water is also 0.0787 mol.

b) Determine the percent yield of H2O for the reaction:

To calculate the percent yield, we need the experimental yield of water, which is given as 3.70 mL.

Mass of water = volume x density = 3.70 mL x 1.00 g/mL = 3.70 g

Number of moles of water = mass of water / molar mass of water = 3.70 g / 18.02 g/mol = 0.2053 mol

Percent yield = (experimental yield / theoretical yield) x 100
Percent yield = (0.2053 mol / 0.0787 mol) x 100 = 261.02%

Therefore, the theoretical yield of H2O for the reaction is 0.0787 mol and the percent yield of H2O for the reaction is 261.02%.

This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.

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

g ethanol = 4.60 mL x 0.789 = approx 3.6 g BUT you need to this calculation and ALL that follow over to obtain a better answer for each step. Mine are just estimates.
mols ethanol = grams/molar mass = approx 0.08
How much H2O would that produce. That's
approx 0.8mols x (3 mols H2O/1 mol ethanol) = apprx 0.08 x 3 = approx 0.24 mols

Now for the O2.
mols O2 = grams/molar mass = approx 0.5
How much H2O would that produce? That's approx 0.5 x 3 mols H2O/3 mols O2) = approx 0.5 mols O2
You see we have two different values for mols H2O and only one can be right. The correct value in LR problems is ALWAYS the smaller value; therefore, we should produce approx 0.24 mol H2O from the ethanol and convert that to grams = approx 4.3 grams. (That's 0.24 mol x 18 g/mol = about 4.3 g). That is the theoretical yield (TY). The actual yield is 3.70 mL which with a density of 1.00 g/mL is 3.70 g and that's the actual yield (AY)
% yield = (AY/TY)*100 = ?