The reaction between 34.0 g of NH3 and excess oxygen gas produces 45.9 g of NO gas and some water. Determine the percent yield.
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To calculate the percent yield, we need to compare the actual yield (the experimental yield obtained) with the theoretical yield (the maximum yield possible given by stoichiometry).
First, let's find the balanced chemical equation for the reaction between NH3 and oxygen gas:
4 NH3 + 5 O2 -> 4 NO + 6 H2O
From the balanced equation, we can see that the molar ratio between NH3 and NO is 4:4 (1:1) and the molar masses are:
NH3 = 17.03 g/mol
NO = 30.01 g/mol
Next, let's calculate the theoretical yield of NO gas:
1 mol of NH3 -> 1 mol of NO
Using the molar masses, we can convert the mass of NH3 to moles:
moles of NH3 = mass of NH3 / molar mass of NH3
moles of NH3 = 34.0 g / 17.03 g/mol
moles of NH3 = 1.995 mol
Since the molar ratio between NH3 and NO is 1:1, the moles of NO gas produced will be the same:
moles of NO = moles of NH3 = 1.995 mol
To find the theoretical yield of NO gas in grams, we can use the molar mass of NO:
mass of NO = moles of NO x molar mass of NO
mass of NO = 1.995 mol x 30.01 g/mol
mass of NO = 59.87 g
Now, let's calculate the percent yield:
percent yield = (actual yield / theoretical yield) x 100
From the given information, the actual yield of NO gas is 45.9 g. Plugging the values into the formula:
percent yield = (45.9 g / 59.87 g) x 100
percent yield = 76.7%
Therefore, the percent yield for the reaction is approximately 76.7%.
To determine the percent yield, we need to compare the actual yield (the amount obtained in the experiment) to the theoretical yield (the amount that would be obtained if the reaction went to completion).
Step 1: Calculate the molar mass of NH3 (ammonia) and NO (nitric oxide).
- The molar mass of NH3: 1(N) + 3(H) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
- The molar mass of NO: 1(N) + 1(O) = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
Step 2: Convert the mass of NH3 to moles.
- Moles NH3 = mass of NH3 / molar mass of NH3
- Moles NH3 = 34.0 g / 17.03 g/mol = 1.995 mol (rounded to 3 decimal places)
Step 3: Determine the stoichiometry of the balanced chemical equation.
- The balanced chemical equation for the reaction is: 4 NH3 + 5 O2 -> 4 NO + 6 H2O
- According to the equation, 4 moles of NH3 react to produce 4 moles of NO.
Step 4: Calculate the theoretical yield of NO.
- Theoretical yield of NO = moles of NH3 * (moles of NO / moles of NH3 in the balanced equation)
- Theoretical yield of NO = 1.995 mol * (4 mol NO / 4 mol NH3) = 1.995 mol
Step 5: Convert the theoretical yield of NO to grams.
- Mass of NO = theoretical yield of NO * molar mass of NO
- Mass of NO = 1.995 mol * 30.01 g/mol = 59.97 g (rounded to 2 decimal places)
Step 6: Calculate the percent yield.
- Percent yield = (actual yield / theoretical yield) * 100%
- Percent yield = (45.9 g / 59.97 g) * 100% = 76.54% (rounded to 2 decimal places)
Therefore, the percent yield of the reaction is approximately 76.54%.
Here is a worked example of a stoichiometry problem that shows how to calculate theoretical yield and percent yield.
http://www.jiskha.com/science/chemistry/stoichiometry.html