What is the percentage yield of N2O5 when 68.0g of O2 reacts With N2 according to the reaction equation, with 72.6g of N2O5 being obtained?
2N2 + 5O2 -> 2N2O5
I assume that is an excess of N2.
2N2 + 5O2 -> 2N2O5
Mols O2 = grams/molar mass = 68.0/32 = approx 2
mols N2O5 formed if 100% yield = approx 2 x (2/5) = about 0,8.
grams N2O5 = mols N2O5 x molar mass N2O4 = ? This is the theoretical yield(TY).
The actual yield (AY) is 72.6 g
% yield = (AY/TY)*100 = ?
Check my work. Remember all of the numbers I've shown are estimates and must be recalculated from scratch.
help
A
Alright, let's break down this problem step by step.
1. Determine the moles of O2:
Moles O2 = mass ÷ molar mass = 68.0 g ÷ 32 g/mol = 2.125 mol
2. Use the mole ratio between O2 and N2O5 to determine the theoretical yield of N2O5:
From the balanced chemical equation, we can see that for every 5 moles of O2, we should get 2 moles of N2O5. Therefore,
Moles N2O5 (from O2) = moles O2 × (2 mol N2O5 ÷ 5 mol O2)
Moles N2O5 = 2.125 mol × (2/5) = 0.85 mol
3. Calculate the theoretical yield of N2O5:
The molar mass of N2O5 is 108 g/mol, so
Theoretical yield = moles N2O5 × molar mass N2O5 = 0.85 mol × 108 g/mol = 91.8 g
4. Calculate the percent yield:
Percent yield = actual yield ÷ theoretical yield × 100%
Percent yield = 72.6 g ÷ 91.8 g × 100% = 79.1%
Therefore, the percentage yield of N2O5 is approximately 79.1%.
Well, let's do some calculations! We start with 68.0g of O2. Now, we need to convert this to moles. Let's use the molar mass of O2, which is 32.0 g/mol. So, 68.0g of O2 is equal to 68.0g / (32.0 g/mol) = 2.125 mol of O2.
According to the balanced equation, we see that the mole ratio between O2 and N2O5 is 5:2. So, if we have 2.125 mol of O2, we can use this ratio to find the moles of N2O5. Let's do the math: 2.125 mol of O2 * (2 mol of N2O5 / 5 mol of O2) = 0.85 mol of N2O5.
Now, let's compare this to the actual amount obtained, which is 72.6g of N2O5. To convert this to moles, we will use the molar mass of N2O5, which is 108.0 g/mol. So, 72.6g of N2O5 is equal to 72.6g / (108.0 g/mol) = 0.67 mol of N2O5.
To calculate the percentage yield, we use the formula: (actual yield / theoretical yield) * 100. In this case, the theoretical yield is 0.85 mol and the actual yield is 0.67 mol. So, the calculation is (0.67 mol / 0.85 mol) * 100 = 78.82%.
Therefore, the percentage yield of N2O5 is approximately 78.82%. However, I must say this reaction seems to be a bit confused. Clearly, it needs some nitrogen to keep it grounded. Otherwise, it's just going to float around like a hot-air balloon. Better N2O5 your surroundings before proceeding!
To find the percentage yield of N2O5, you need to compare the actual yield (72.6g) with the theoretical yield.
Step 1: Determine the moles of O2 used and N2O5 obtained.
To find the moles of O2 used, divide the given mass of O2 by its molar mass.
Molar mass of O2 = 32.00 g/mol
Moles of O2 = 68.0 g / 32.00 g/mol = 2.125 mol
To find the moles of N2O5 obtained, divide the given mass of N2O5 by its molar mass.
Molar mass of N2O5 = 108.02 g/mol
Moles of N2O5 = 72.6 g / 108.02 g/mol = 0.672 mol
Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, you can see that 2 moles of N2 react with 5 moles of O2 to produce 2 moles of N2O5.
Step 3: Calculate the theoretical yield of N2O5.
Using the moles of O2 (2.125 mol) as the limiting reactant, you can calculate the moles of N2O5 that should be produced.
According to the stoichiometry, for every 5 moles of O2, 2 moles of N2O5 are produced.
Moles of N2O5 (theoretical) = (2.125 mol O2) x (2 mol N2O5 / 5 mol O2) = 0.85 mol
Step 4: Calculate the percentage yield.
The percentage yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100%.
Percentage yield = (actual yield / theoretical yield) x 100%
Percentage yield = (0.672 mol / 0.85 mol) x 100% = 79.06%
So, the percentage yield of N2O5 in this reaction is approximately 79.06%.