if 15.5 g of dinitrogen dioxide 600.1 g is produced, what is percent yield
To calculate the percent yield, we need to compare the actual yield (the amount actually obtained) to the theoretical yield (the maximum possible amount based on stoichiometry).
First, we need to determine the balanced equation for the reaction involving dinitrogen dioxide. Let's assume the reaction is:
2 NO2(g) → N2O4(g)
The molar mass of NO2 is 46.01 g/mol, so we can calculate the number of moles of NO2 produced using the given mass:
moles of NO2 = mass of NO2 / molar mass of NO2
moles of NO2 = 15.5 g / 46.01 g/mol
moles of NO2 = 0.3369 mol
According to the balanced equation, the ratio of moles of NO2 to moles of N2O4 is 2:1. So, the theoretical yield of N2O4 can be calculated:
moles of N2O4 (theoretical) = 0.3369 mol NO2 × (1 mol N2O4 / 2 mol NO2)
moles of N2O4 (theoretical) = 0.3369 mol / 2
moles of N2O4 (theoretical) = 0.1685 mol
Now, let's calculate the actual yield of N2O4:
mass of N2O4 (actual) = 600.1 g
To convert this mass into moles, we need the molar mass of N2O4. The molar mass of N2O4 is 92.011 g/mol.
moles of N2O4 (actual) = mass of N2O4 (actual) / molar mass of N2O4
moles of N2O4 (actual) = 600.1 g / 92.011 g/mol
moles of N2O4 (actual) = 6.5236 mol
Finally, we can calculate the percent yield using the formula:
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (6.5236 mol / 0.1685 mol) × 100
Percent yield = 38.7%
Therefore, the percent yield of N2O4 in this reaction is approximately 38.7%.