2n2o5(g)→4no2(g)+o2(g) when 40g of n2o5 decompose 4.5g of o2 is formed what is the percent yield
the equation says you should get 1/2 mole of O2 for every mole of N2O5
40g = 0.37 moles N2O5
4.5g = 0.14 moles O2
0.14/0.185 = 0.756 = 75.6% yield
Using the balanced equation:
2N2O5 (g) -> 4NO2 (g) + O2 (g)
We can find the theoretical yield of O2 produced by calculating the amount of N2O5 consumed, and then using the mole ratio to find the corresponding amount of O2 produced.
40 g of N2O5 is equal to 0.37 moles of N2O5 (molar mass of N2O5 is 108 g/mol).
According to the equation, for every 2 moles of N2O5 that decompose, 1 mole of O2 is produced. So, the theoretical yield of O2 is:
(0.37 mol N2O5 / 2 mol N2O5) x 1 mol O2 = 0.19 mol O2
Next, we can calculate the percent yield by comparing the actual yield of O2 (4.5 g) to the theoretical yield we just calculated (0.19 mol), and converting to percent:
Actual yield: 4.5 g O2 x (1 mol O2 / 32 g O2) = 0.14 mol O2
Percent yield = (actual yield / theoretical yield) x 100%
Percent yield = (0.14 / 0.19) x 100% = 73.7%
Therefore, the percent yield of the reaction is approximately 73.7%
N2O5 /4NO2+O2?
This is a chemical equation that represents the decomposition of nitrogen pentoxide (N2O5) into nitrogen dioxide (NO2) and oxygen gas (O2). The balanced equation is:
2N2O5 (g) -> 4NO2 (g) + O2 (g)
This means that for every two moles of N2O5 that decompose, four moles of NO2 and one mole of O2 are produced.
When 40g of N2O5 decoposes 4.5 g of O2 formed what is present yieud
Well, let's calculate the theoretical yield first. From the balanced equation, we can see that 2 moles of N2O5 produce 1 mole of O2. So, we need to convert the grams of N2O5 to moles first.
Given that the molar mass of N2O5 is approximately 108 grams/mol:
40g N2O5 * (1 mol/108g) = 0.37 mol N2O5
Since 2 moles of N2O5 produce 1 mole of O2:
0.37 mol N2O5 * (1 mol O2 / 2 mol N2O5) = 0.185 mol O2 (theoretical yield)
Now, let's calculate the percent yield. We are given that 4.5g of O2 is actually formed. So, we need to convert grams of O2 to moles.
Given that the molar mass of O2 is approximately 32 grams/mol:
4.5g O2 * (1 mol/32g) = 0.14 mol O2 (actual yield)
Now we can calculate the percent yield:
Percent yield = [Actual yield (mol) / Theoretical yield (mol)] * 100
Percent yield = (0.14 mol O2 / 0.185 mol O2) * 100
Percent yield = 75.7%
So, the percent yield is approximately 75.7%. However, since I'm Clown Bot, I'd like to tell you that chemistry anxiety can make you lose your electrons, so make sure to keep an ion them!
To calculate the percent yield, you need to compare the actual yield (the amount of product that is obtained in the experiment) to the theoretical yield (the amount of product that should ideally be obtained according to the balanced equation).
In this reaction, the balanced equation shows that 2 moles of N2O5 produces 1 mole of O2. Therefore, you need to determine the number of moles of N2O5 to find the theoretical yield of O2.
1. Calculate the number of moles of N2O5:
- Given mass of N2O5 = 40g
- Molar mass of N2O5 = 2(14.01) + 5(16.00) = 108.01 g/mol
- Moles of N2O5 = mass / molar mass = 40g / 108.01 g/mol
2. Determine the theoretical yield of O2:
- The molar ratio between N2O5 and O2 is 2:1, so the moles of O2 will be half of the moles of N2O5.
- The theoretical yield of O2 = Moles of N2O5 / 2
3. Calculate the actual yield of O2:
- Given mass of O2 = 4.5g
4. Calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) x 100
Now, you can substitute the values into the formula to calculate the percent yield.