Question

Calculate the volume of nitrogen dioxide produced at 798.6 torr and 21.1°C by the reaction of 8.75 cm³ copper (density = 8.95 g/cm³) with 228.6 mL of concentrated nitric acid if the acid has a density of 1.42 g/cm³and contains 68.0% HNO₃ by mass).
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Answer 1

mass of Cu = 8.75 x 8.95 = 78.3125 g

moles of Cu = 78.3125 / 63.55 = 1.232

mass of HNO3 = 228.6 x 1.42 = 324.6 g

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

68 = mass of HNO3 / 324.6 ) x 100

mass of HNO3 = 220.728 g

moles of HNO3 = 3.503 mol

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

   1            4                                                                   2

1.232      3.503                                                               ??

here limiting reagent is HNO3.

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

3.503 mol HNO3   -------------> ??

moles of NO2 = 3.503 x 2 / 4 = 1.75

pressure = 798.6 / 760 = 1.051 atm

temperature = 21.1 oC = 294.25 K

P V = n R T

1.051 x V = 1.75 x 0.0821 x 294.25

V = 40.22 L

volume of NO2 = 40.2 L