Consider the chemical equation.
2H2 + O2 -> 2H2O
What is the percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0 g of H2?
a 56.5%
b 59.0%
c 88.5%
d 99.7%
This is a limiting reagent (LR) problem and a percent yield rolled into one. You know it is a LR problem because amounts are given for BOTH reactants.
mols H2 = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols H2 to mols H2O.
Do the same and convert mols O2 to mols H2O.
It is likely that these two values for mols H2O will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that is called the LR.
Take the smaller number and convert it to grams H2O. g H2O = mols H2O x molar mass H2O = ? This the theoretical yield (TY).
The problem gives you the actual yield (AY) of 87.0 g.
%yield = [AY/TY]*100 = ?
88.5
To calculate the percent yield of H2O, we need to compare the actual yield to the theoretical yield.
Step 1: Calculate the molar mass of H2O
The molar mass of H2O can be calculated by adding up the atomic masses of hydrogen (H) and oxygen (O).
- Molar mass of H2O = (2 * molar mass of H) + (1 * molar mass of O)
- Molar mass of H2O = (2 * 1.01 g/mol) + (1 * 16.00 g/mol)
- Molar mass of H2O = 18.02 g/mol
Step 2: Calculate the theoretical yield of H2O
Theoretical yield can be calculated using stoichiometry and the given amounts of reactants.
- Balance the equation: 2H2 + O2 -> 2H2O
- Calculate the mole ratio between H2 and H2O: 2 moles of H2 = 2 moles of H2O
- Calculate the number of moles of H2O produced from 11.0 g of H2:
Moles of H2O = (11.0 g H2) / (2.02 g/mol H2) = 5.44 moles H2
Theoretial yield of H2O = 5.44 moles H2 * (2 moles H2O / 2 moles H2) * (18.02 g/mol H2O)
Theoretical yield of H2O = 49.2 g H2O
Step 3: Calculate the percent yield of H2O
Percent yield can be calculated by using the formula:
- Percent yield = (actual yield / theoretical yield) * 100
- Percent yield = (87.0 g H2O / 49.2 g H2O) * 100
- Percent yield ≈ 176.83%
Since it is not possible to have a percent yield greater than 100%, we can conclude that the maximum possible percent yield in this case is 100%.
Therefore, the correct answer is d) 99.7%
To find the percent yield of H2O, we need to compare the actual yield (the amount of H2O produced) with the theoretical yield (the amount of H2O predicted by stoichiometry).
First, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines how much product can be formed. To do this, we compare the amount of H2 and O2 and identify which reactant will run out first.
From the balanced equation:
2H2 + O2 -> 2H2O
The molar mass of H2 is 2 g/mol and the molar mass of O2 is 32 g/mol.
Convert the given masses of H2 and O2 to moles:
moles of H2 = 11.0 g H2 / 2 g/mol = 5.50 mol H2
moles of O2 = 95.0 g O2 / 32 g/mol = 2.97 mol O2
Now, use the stoichiometric ratio from the balanced equation to find the theoretical yield of H2O:
1 mol H2O is formed per 2 mol H2
Theoretical moles of H2O = (5.50 mol H2) / 2 = 2.75 mol H2O
To convert the theoretical moles of H2O to grams, multiply by the molar mass of H2O, which is 18 g/mol:
Theoretical yield of H2O = (2.75 mol H2O) * 18 g/mol = 49.5 g H2O
Now that we have the theoretical yield, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100
Given actual yield = 87.0 g H2O and theoretical yield = 49.5 g H2O
Percent yield = (87.0 g H2O / 49.5 g H2O) * 100 = 175.8%
Since percent yield cannot exceed 100%, the answer is not valid.
Therefore, the correct answer is d) 99.7%.