Question

Calculate the volume of nitrogen dioxide produced at 712.5 torr and 23.3°C by the reaction of 7.85 cm3 copper (density = 8.95 g/cm3) with 207.3 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

Solution : -

Balanced reaction equation

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

Lets first calculate the mass of the HNO3 solution

Mass = volume * density

          = 207.3 ml * 1.42 g/ml

          = 294.4 g

Now lets calculate the mass of HNO3 in the solution

294.4 g solution * 68 % HNO3 / 100 %= 200 g HNO3

Lets calculate the mass of copper

Mass of copper = 7.85 cm3*8.95 g/cm3 = 70.3 g Cu

Now lets calculate the moles of the Cu and HNO3

Moles = mass / molar mass

Moles of Cu = 70.3 g Cu / 63.546 g per mol = 1.11 mol Cu

Moles of HNO3 = 200 g /63.012 g per mol = 3.174 mol HNO3

Mole ratio of the Cu to HNO3 is 1 : 4

So lets calculate the moles of Cu needed to react with 3.174 mol HNO3

3.174 mol HNO3 * 1 mol Cu / 4 mol HNO3 = 0.7935 mol Cu

Moles of Cu present are more than needed for the reaction therefore HNO3 is the limiting reactant

Now lets calculate the moles of the NO2 gas produced using the mole ratio of the HNO3

3.174 mol HNO3 * 2 mol NO2 / 4 mol HNO3 = 1.587 mol NO2

Now lets calculate the volume of the NO2 at the given conditions

Temperature T= 23.3 C +273 = 296.3 K

Pressure P = 712.5 torr * 1 atm / 760 torr = 0.9375 atm

Ideal gas equation

PV= nRT

Therefore

V= nRT/ P

R= 0.08206 L atm per mol K

Lets put the values in the formula

V= 1.587 mol * 0.08206 L atm per mol K * 296.3 K / 0.9375 atm

V= 41.2 L

Therefore volume of the NO2 gas produced = 41.2 L