For the following reaction, 5.70 grams of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 22.6 grams of ammonia.
(1) What is the theoretical yield of ammonia ? ____ grams
(2) What is the percent yield for this reaction ? ____ %
(1) What is the theoretical yield of ammonia? ____ grams
Well, the reaction you described is like a magic trick - turning hydrogen and nitrogen into ammonia! It's almost like turning water into wine. But instead of wine, we get ammonia. Clever, huh? Now, to answer your question, we need to do a little math.
First, we convert the mass of hydrogen gas to moles. Hydrogen has a molar mass of around 2 grams per mole, so 5.70 grams of hydrogen would be approximately 2.85 moles.
The balanced equation for the reaction is N2 + 3H2 → 2NH3.
From this equation, we can see that for every two moles of ammonia produced, we need three moles of hydrogen gas. So, using this ratio, we can determine the number of moles of ammonia formed.
Since we have 2.85 moles of hydrogen, we can set up a proportion:
(2 mol NH3 / 3 mol H2) = (x moles NH3 / 2.85 moles H2)
By cross-multiplying and solving for 'x', we find that x = 1.9 moles of NH3.
Now, to find the theoretical yield of ammonia in grams, we multiply the number of moles by the molar mass of ammonia, which is about 17 grams per mole:
1.9 moles NH3 × 17 g/mol NH3 = ~32.3 grams of ammonia.
So, the theoretical yield of ammonia is approximately 32.3 grams!
(2) What is the percent yield for this reaction? ____ %
The percent yield is a measure of how efficient the reaction is in producing the desired product. It's like being graded on how well you did your magic trick!
To calculate the percent yield, we need to compare the actual yield (22.6 grams) to the theoretical yield (32.3 grams) we calculated earlier.
Percent Yield = (Actual Yield / Theoretical Yield) × 100%
Plugging in the values, we get:
(22.6 g NH3 / 32.3 g NH3) × 100% = ~70%
So, the percent yield for this reaction is approximately 70%. Not bad for a little magic trick, huh?
To find the theoretical yield of ammonia in this reaction, you need to calculate how much ammonia would be produced if all the reactants were converted to product according to the balanced chemical equation.
The balanced chemical equation for this reaction is:
N2 + 3H2 -> 2NH3
First, convert the given mass of hydrogen gas to moles. To do this, divide the given mass by the molar mass of hydrogen gas, which is 2.016 grams/mol.
5.70 grams H2 / 2.016 grams/mol = 2.83 mol H2
From the balanced equation, you can see that the mole ratio between hydrogen gas and ammonia is 3:2. This means that for every 3 moles of hydrogen gas reacted, 2 moles of ammonia are produced.
Using this ratio, you can calculate the moles of ammonia produced:
(2.83 mol H2) * (2 mol NH3 / 3 mol H2) = 1.89 mol NH3
Now, you need to convert the moles of ammonia to grams. To do this, multiply the moles of ammonia by the molar mass of ammonia, which is 17.031 grams/mol.
(1.89 mol NH3) * (17.031 grams/mol NH3) = 32.19 grams NH3
Therefore, the theoretical yield of ammonia in this reaction is 32.19 grams.
To calculate the percent yield, divide the actual yield (given as 22.6 grams) by the theoretical yield (calculated as 32.19 grams) and multiply by 100:
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (22.6 grams / 32.19 grams) * 100
Percent yield = 70.24%
Therefore, the percent yield for this reaction is 70.24%.