When NH3 is prepared from 28 g of N2 and excess H2, the theoretical yield of NH3 is 34 g. When this reaction is carried out in a given experiment, only 30 g is produced. What is the percent yield?
% yield = (actual/theoretical)*100 =
(30/34)*100 = ??
Well, it seems like the reaction couldn't quite measure up to expectations! Let's calculate the percent yield to see just how close it got.
The theoretical yield is given as 34 g, while the actual yield is 30 g.
To calculate the percent yield, we use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
So, plugging in our values:
Percent Yield = (30 g / 34 g) x 100
Hmmm, let me do some math here.
Percent Yield ≈ 88.2%
Well, it looks like the reaction managed to achieve about 88.2% of the expected yield. Not too shabby, but certainly not perfect! Maybe it just needs a little extra encouragement next time. Keep trying, reaction!
To find the percent yield, you can use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Given that the actual yield is 30 g and the theoretical yield is 34 g, we can substitute these values into the formula to calculate the percent yield:
Percent Yield = (30 g / 34 g) * 100
Calculating this expression gives us:
Percent Yield = 0.8824 * 100
So the percent yield is approximately 88.24%.
To calculate the percent yield, you need to compare the actual yield to the theoretical yield and express it as a percentage.
First, let's determine the theoretical yield:
The balanced equation for the reaction is:
N2 + 3H2 -> 2NH3
From the reaction equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
To find the number of moles of N2:
molar mass of N2 = 2(atomic mass of nitrogen)
molar mass of N2 = 2(14.01 g/mol) = 28.02 g/mol
number of moles of N2 = mass of N2 / molar mass of N2
number of moles of N2 = 28 g / 28.02 g/mol
number of moles of N2 ≈ 1 mol
Since 1 mole of N2 reacts to produce 2 moles of NH3, the theoretical yield of NH3 would be:
2 moles NH3 * molar mass of NH3
The molar mass of NH3 = (1)(atomic mass of nitrogen) + (3)(atomic mass of hydrogen)
The molar mass of NH3 = (1)(14.01 g/mol) + (3)(1.008 g/mol)
The molar mass of NH3 = 14.01 g/mol + 3.024 g/mol
The molar mass of NH3 ≈ 17.034 g/mol
theoretical yield of NH3 = 2 moles * 17.034 g/mol ≈ 34.068 g
Now, let's calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100
percent yield = (30 g / 34.068 g) * 100
percent yield ≈ 88.04%
Therefore, the percent yield of NH3 in this experiment is approximately 88.04%.