Question

The following reaction forms 15.8 g of Ag:

2 Ag2O---> 4 Ag + O2

What total volume of gas forms if it is collected over water at a temperature of 25 and a total pressure of 752mmHg?

Answer 1

Answer – We are given, mass of Ag = 15.8 g , T = 25 +273 = 298 K

Total pressure = 752 mm Hg

First we are calculating the moles of 15.8 g

Moles of Ag = 15.8 g / 107.87 g.mol^-1

                      = 0.146 moles

From the balanced reaction

4 moles of Ag = 1 moles of O₂

So, 0.146 moles of Ag = ?

= 0.0366 moles O₂

Now we have T, P and n and from that we need to calculate V

Pressure of water vapour at 25^oC = 23.8 mm Hg

So pressure of gas = 752 – 23.8 = 728.2

P = 728.2 mm Hg * 1 atm/ 760mm Hg

    = 0.958 atm

Using the Ideal gas law

PV = nRT

So, V = nRT/P

          = 0.0366 moles * 0.0821 L.atm.mol^-1.K^-1 *298 K / 0.958 atm

          = 0.935 L

          = 935 mL

So, total volume of gas forms is 935 mL or 0.935 L