Question

Oxygen gas, generated by the reaction 2 KClO 3 ---- 2 KCl + 3 O 2 , is collected over water at 27 C in a 2.00 L vessel at a total pressure of 760. mmHg. The vapor pressure of H 2 O at 27 C is 26 mmHg. How many moles of KClO 3 were consumed in the reaction?

Answer 1

Answer:

The moles of KClO3 = 0.052 moles

Explanation:

Step 1: Calculate the pressure of oxygen gas

The oxygen has a total pressure (including water vapour) of 760 mmHg

The pressure of Oxygen = (760 - 26) mmHg

                                         = 734 mmHg of water vapor

Step 2: Calculate the no of moles of oxygen

        Using Ideal gas equation

        P V = n R T

       P = pressure of oxygen in N/m2 ( you should convert 734 mmHg to pascal or N/m2) = 97,858.6 N/m2 or pas

       V = 2 litres = 0.002 m3

       R = gas constant = 8.31

       T= 27oC = 300 K

Applying this equation P V = n R T

        97,858.6 x 0.002 = n x 8.31 x 300

        n = 0.0785 mol of Oxygen

From the balanced equation

2 KClO 3 ---- 2 KCl + 3 O 2

3 moles of oxygen is produced from 2 moles KClO3

so 0.0785 mole of oxygen will be produced from x

x = (0.0785 x 2 ) / 3

x = 0.052 moles of KClO3