The experimental yield of nitric acid is 24.648 g if percent yield for the reaction is 60%.
Nitrogen dioxide mass, given, is 45.0 g
Nitrogen dioxide has a molecular mass of 46 g/mol.
For the chemical formula given:
Nitrogen dioxide moles are equal to 45/46 or 0.978 g/mol.
Using the reaction's stoichiometry:
2 moles of nitric acid are produced from 3 moles of nitrogen dioxide.
Nitric acid will therefore be produced by 0.978 moles of nitrogen dioxide.
Now, using equation 1 to calculate the mass of nitric acid, we obtain:
Nitric acid's molecular weight is 63 g/mol.
Nitric acid equals 0.652 moles = weight of nitric acid/ 63
weight of nitric acid= 41.076g
Equation 1 is solved for the following values:
We use the equation: to determine the experimental yield of nitric acid.
0.652=experimental yield / theoretical yield * 100
Theoretical yield = 41.08 g
60%= experimemtal yield/ 41.08*100
Experimental yield= 24.648 g
For more information on experimental yield kindly visit to
https://brainly.com/question/17645320
#SPJ4
