Question

A student heats a sample of KClO3, KCl, and MnO2 having a mass of 3.63.49 g. After the reaction is complete, she has collected 346.2 mL of oxygen in a water filled flask. The room temperature is 23.6 degrees C and the atmospheric pressure is 763.4 mmHg. The residue had a mass of 3.1915 g.

a. What is the partial pressure of the oxygen? The vapor pressure of water is 21.3 mmHg at 23.6 degrees C.
b. Also what mass of oxygen was liberated?
c. And how many moles of oxygen were produced?

Answer 1

a) 742.1 mm Hg O2

b) 0.4434 g O2

c) 0.01386 mol O2

Step-by-step explanation:

Step 1: Data given

Mass of the sample = 3.6349 grams

volume of oxygen = 346.2 mL

Temperature = 23.6 °C

atmospheric pressure = 763.4 mmHg

Step 2: What is the partial pressure of the oxygen?

763.4 mmHg - 21.3 mmHg = 742.1 mm Hg O2

b. What mass of oxygen was liberated?

All of the mass lost by the reactants was due to oxygen being liberated.

(3.6349 g) - ( 3.1915 g) = 0.4434 g O2

c. How many moles of oxygen were produced?

(0.4434 g O2) / (32 g O2/mol) = 0.01386 mol O2