15.8 grams of NH3 and excess O2 produces NO gas and H2O. 12.8 grams of NO gas was produced. Find the percent yield.
Balanced Equation: 4 NH3 + 5 O2 -> 4 NO + 6 H2O
Here is an example of a simple stoichiometry problem. Percent yield is given at the bottom of the example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the percent yield, we need to calculate the theoretical yield and the actual yield, and then use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
First, let's calculate the theoretical yield of NO gas using the given mass of NH3.
Step 1: Convert the mass of NH3 to moles.
Using the molar mass of NH3 (17.03 g/mol):
Number of moles of NH3 = Mass of NH3 / Molar mass of NH3 = 15.8 g / 17.03 g/mol ≈ 0.9277 mol NH3
Step 2: Calculate the mole ratio between NH3 and NO.
From the balanced equation, we see that the mole ratio between NH3 and NO is 4:4.
Therefore, the number of moles of NO produced will be the same as the number of moles of NH3.
Step 3: Convert the moles of NO to grams.
Using the molar mass of NO (30.01 g/mol):
Theoretical yield of NO = Number of moles of NO x Molar mass of NO = 0.9277 mol x 30.01 g/mol ≈ 27.839 g
Now let's calculate the actual yield of NO gas using the given mass of NO.
Actual Yield of NO = 12.8 g
Now we can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Percent Yield = (12.8 g / 27.839 g) x 100 ≈ 45.96%
Therefore, the percent yield is approximately 45.96%.