Nitrogen dioxide (NO2) gas and liquid water (H2O) react to form aqueous nitric acid (HNO3) and nitrogen monoxide (NO) gas. Suppose you have 5.0 mol of
NO2 and 11.0 mol of H2O in a reactor.
Suppose as much as possible of the NO2 reacts. How much will be left? Round your answer to the nearest 0.1 mol.

9.3 moles of H20

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

3NO2 + H2O —> 2HNO3 + NO

Data obtained from the question include:

Number of mole of NO2 = 5moles

Number of mole of H2O = 11moles

From the balanced equation,

3 moles of NO2 reacted with 1 mole of H2O.

Therefore, 5 moles of NO2 will react with = 5/3 = 1.7 mole of H2O

From the calculation above, it is obvious that some amount of H20 did not react.

The unreacted H2O = 11 —1.7 = 9.3 moles

Therefore, 9.3 moles of H20 is left over

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole