If the percent yield for the following reaction is 80.37%, and 63.21 g of no2 are used in the reaction mixture, how many grams of nitric acid, hno3(aq), are produced in the experiment? 3 no2(g) + h2o(l) → 2 hno3(aq) + no(g) molar mass no2 = 46.01 g/mol and hno3 = 63.01 g/mol.

Explanation:

It is given that mass of is 46.01 g/mol and mass of is 63.01 g/mol.

So, in 1 mole calculate the number of moles present in 63.21 g/mol of as follows.

                 No. of moles =

                                       =

                                       = 1.37 mol

Now, according to the reaction equation we require 3 moles of to make 2 moles of .

                          = 0.91 mol

Hence, calculate the mass of as follows.

                  No. of moles =

                           0.91 mol =

                          mass = 57.33 g

Now, mass of actually produced by 80.37% yield is calculated as follows.

                       = 46.07 g

Thus, we can conclude that 46.07 g of nitric acid, , are produced in the experiment.

increased stability in the completion of fluorine's outer shell.

i think   the answer is c hope i'm right