What mass of N2O5 will result from the reaction of 6.0 mol of NO2 if there is a 61.2% yield in the reaction?
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Reaction NO2 with what. O2?
4NO2 + O2 ==> 2N2O5
mols NO2 = 6.0
How many mols N2O5 will result?
That's 6.0 mols NO2 x (2 mols N2O5/4 mols NO2) = 6.0 x (2/4) = 3.0 mols N2O5.
g N2O5 = mols x molar mass = ? if the yield was 100%.
If 61.2% yield, then ? x 0.612 = x.
To find the mass of N2O5, we need to use the given information about the yield of the reaction.
1. Calculate the moles of N2O5 produced:
Since the reaction is given as the reaction of 6.0 mol of NO2, we need to determine the balanced equation for the reaction to find the stoichiometric ratio between NO2 and N2O5.
2NO2 + O2 -> 2N2O5
From the equation, we can see that 2 moles of NO2 react to form 2 moles of N2O5. So, the ratio is 2:2 or 1:1.
Since we have 6.0 mol of NO2, the moles of N2O5 formed will also be 6.0 mol.
2. Calculate the actual yield of N2O5:
The actual yield is given as 61.2% of the theoretical yield. To calculate the actual yield, we need to multiply the theoretical yield by the percentage yield.
Actual yield = 61.2% × theoretical yield
= 0.612 × 6.0 mol
= 3.672 mol
3. Calculate the molar mass of N2O5:
The molar mass of N2O5 is calculated by summing the atomic masses of nitrogen (N) and oxygen (O).
Molar mass of N2O5 = (2 × atomic mass of N) + (5 × atomic mass of O)
= (2 × 14.01 g/mol) + (5 × 16.00 g/mol)
= 28.02 g/mol + 80.00 g/mol
= 108.02 g/mol
4. Calculate the mass of N2O5:
To find the mass, we multiply the molar mass of N2O5 by the number of moles.
Mass of N2O5 = molar mass × moles
= 108.02 g/mol × 3.672 mol
= 396.40744 g
Therefore, the mass of N2O5 that will result from the reaction of 6.0 mol of NO2 with a 61.2% yield is approximately 396.41 g.
To calculate the mass of N2O5 produced, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction.
The balanced equation for the reaction is:
2NO2 + O2 -> 2N2O5
This equation tells us that two moles of NO2 react to form two moles of N2O5.
Step 2: Determine the limiting reagent.
The limiting reagent is the reactant that is completely consumed in the reaction, thus determining the maximum amount of product that can be formed.
In this case, we have 6.0 mol of NO2, and according to the balanced equation, 2 moles of NO2 are required to produce 2 moles of N2O5. Therefore, the stoichiometric ratio between NO2 and N2O5 is 1:1.
Since the stoichiometry is 1:1, the limiting reagent is NO2.
Step 3: Calculate the theoretical yield.
The theoretical yield is the maximum amount of product that can be formed based on the stoichiometry of the reaction.
Since we have determined that NO2 is the limiting reagent, the amount of N2O5 produced will be equal to the amount of NO2 used. Therefore, the theoretical yield of N2O5 is also 6.0 mol.
Step 4: Calculate the actual yield.
The actual yield is the amount of product obtained experimentally. In this case, the problem states that the yield in the reaction is 61.2%. To calculate the actual yield, we multiply the theoretical yield by the yield percentage:
Actual yield = Theoretical yield × Yield percentage
Actual yield = 6.0 mol × 0.612
Actual yield = 3.672 mol
Step 5: Calculate the mass of N2O5 produced.
To calculate the mass of N2O5 produced, we need to use the molar mass of N2O5, which is 108.02 g/mol.
Mass of N2O5 = Actual yield × Molar mass of N2O5
Mass of N2O5 = 3.672 mol × 108.02 g/mol
Mass of N2O5 = 396.99 g
Therefore, the mass of N2O5 that will result from the reaction of 6.0 mol of NO2 with a 61.2% yield is 396.99 g.