When 100.g Mg3N2 reacts with 75.0 g H2O, 15.0 g of NH3 is formed. What is the % yield?
% yield ... 100*[15.0 g / (1.39 mol * 17.0 g/mol)]
Well, when it comes to percentage yield, we need to bring out our mathematical hats. And maybe put on some clown makeup while we're at it! So, let's do some number crunching.
Step 1: Calculate the molar mass of NH3 (ammonia). Nitrogen (N) has a molar mass of approximately 14.01 g/mol, and since there are three hydrogens (H) in ammonia, each with a molar mass of approximately 1.01 g/mol, the molar mass of ammonia (NH3) is around 17.03 g/mol.
Step 2: Now let's find out how many moles of NH3 you should have obtained. By dividing the mass of NH3 formed (15.0 g) by the molar mass of NH3 (17.03 g/mol), we get approximately 0.88 moles.
Step 3: Determine the theoretical yield of NH3, assuming the reaction goes perfectly. From the balanced chemical equation, we can see that 1 mole of Mg3N2 produces 2 moles of NH3. So if we start with 0.88 moles of Mg3N2, we should theoretically get 1.76 moles of NH3.
Step 4: Lastly, let's calculate the percentage yield. Divide the actual yield (0.88 moles) by the theoretical yield (1.76 moles) and multiply by 100 to get the answer in percent. So, the percentage yield would be around 50%.
But hey, remember that chemical reactions aren't always perfect, just like a clown's attempt at juggling. So, sometimes the actual yield can be less than the theoretical yield. Don't worry, though! Practice makes perfect, and clowns are all about putting on a show!
To calculate the percent yield, we need to compare the actual yield of the product with the theoretical yield. The theoretical yield is the amount of product that would be obtained if the reaction went to completion with 100% efficiency. The actual yield, on the other hand, is the amount of product obtained in a real experiment.
First, let's calculate the theoretical yield of NH3. We'll start by finding the limiting reactant, which is the reactant that is completely consumed in the reaction. The balanced chemical equation for the reaction is:
Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3
The molar mass of Mg3N2 is 100.g/mol.
The molar mass of H2O is 18.015 g/mol.
The molar mass of NH3 is 17.031 g/mol.
To find the limiting reactant, we need to calculate how many moles of each reactant we have:
Number of moles of Mg3N2 = mass of Mg3N2 / molar mass of Mg3N2
= 100.g / 100.g/mol
= 1 mol
Number of moles of H2O = mass of H2O / molar mass of H2O
= 75.0 g / 18.015 g/mol
= 4.162 mol
According to the balanced chemical equation, each mole of Mg3N2 reacts to produce 2 moles of NH3. Therefore, the maximum number of moles of NH3 that can be produced is:
Maximum moles of NH3 = 2 * number of moles of Mg3N2
= 2 * 1 mol
= 2 mol
Now, let's calculate the theoretical yield of NH3 in grams:
Theoretical yield of NH3 = maximum moles of NH3 * molar mass of NH3
= 2 mol * 17.031 g/mol
= 34.062 g
The actual yield of NH3 is given in the question as 15.0 g.
Finally, let's calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100%
= (15.0 g / 34.062 g) * 100%
= 44.01%
Therefore, the percent yield of NH3 is approximately 44.01%.
To calculate the percent yield, we need to compare the actual yield with the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained based on stoichiometry and assuming 100% reaction efficiency. The actual yield is the amount of product obtained experimentally.
First, we need to determine the balanced chemical equation for the reaction between Mg3N2 and H2O:
Mg3N2 + 6 H2O -> 3 Mg(OH)2 + 2 NH3
From the balanced equation, we can see that 1 mole of Mg3N2 reacts to give 2 moles of NH3. Therefore, we need to convert the given masses of Mg3N2 and NH3 to moles.
The molar mass of Mg3N2 is 100 g/mol, so the number of moles of Mg3N2 is calculated as:
moles of Mg3N2 = mass of Mg3N2 / molar mass of Mg3N2
= 100 g / 100 g/mol
= 1 mol
In a similar way, the molar mass of NH3 is 17 g/mol, so the number of moles of NH3 formed in the reaction is:
moles of NH3 = mass of NH3 / molar mass of NH3
= 15.0 g / 17 g/mol
= 0.882 mol
Now, we can calculate the theoretical yield of NH3. Since the reaction has a 1:2 stoichiometric ratio between Mg3N2 and NH3, we can use the number of moles of Mg3N2 to determine the theoretical yield.
theoretical yield of NH3 = moles of Mg3N2 × (moles of NH3 / moles of Mg3N2)
= 1 mol × (2 mol / 1 mol)
= 2 mol
Thus, the theoretical yield of NH3 is 2 moles.
Finally, we can calculate the percent yield using the actual yield and theoretical yield:
percent yield = (actual yield / theoretical yield) × 100
= (0.882 mol / 2 mol) × 100
= 44.1%
Therefore, the percent yield of NH3 in this reaction is 44.1%.
Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g)
water is the limiting reactant ... 75.0 g / 18.0 g/mole = 4.17 moles
expected NH3 ... 4.17 / 3 = 1.39 moles
... molar mass NH3 ... 17.0 g
% yield ... 15.0 g / (1.39 mol * 17.0 g/mol)