Question

Calculate the volume of nitrogen dioxide produced at 794.8 torr and 25.6°C by the reaction of 9.75 cm³ copper (density = 8.95 g/cm³) with 207.2 mL of concentrated nitric acid if the acid has a density of 1.42 g/cm³ and contains 68.0% HNO₃ by mass).
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Answer 1

mass of Cu = 9.75 x 8.95 = 87.2625 g

moles of Cu = 87.2625 / 63.55 = 1.373

mass of HNO3 = 207.2 x 1.42 = 294.224 g

mass % = mass of HNO3 / mass of solution ) x 100

68 = mass of HNO3 / 294.224 ) x 100

mass of HNO3 = 200.07 g

moles of HNO3 = 3.175 mol

Cu(s) + 4HNO₃(aq) -----------------> Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

   1            4                                                                   2

1.373      3.175                                                               ??

here limiting reagent is HNO3.

4 mol HNO3 -------------> 2mol NO2

3.175 mol HNO3   -------------> ??

moles of NO2 = 3.175 x 2 / 4 = 1.588

pressure = 798.6 / 760 = 1.046 atm

temperature = 21.1 oC = 298.75 K

P V = n R T

1.046 x V = 1.588 x 0.0821 x 298.75

V = 37.24 L

volume of NO2 = 37.24 L