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
2n2o5(g)→4no2(g)+o2(g) when 40g of n2o5 decompose 4.5g of o2 is formed what is the percent yield
5 answers
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.
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
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%
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%
3 answers