Gas volume answer 3a

Potassium chlorate(V) decomposes on heating to produce oxygen gas. It is suggested that the equation for the reaction is one of the following two:

KClO₃(s) ⇒ KClO + O₂(g)

KClO₃(s) ⇒ KCl + 1½O₂(g)

Plan a gas volume experiment to determine which is the correct equation.

We shall use the same side arm test tube and gas syringe apparatus which was used in exercise 3. We shall aim to have no more than 60 cm³ of gas (based on the second equation as this gives the greatest volume of gas). The decomposition doesn't take place until around 600 ºC and so we need to allow for expansion of the gases in the test tube. If we produce too much gas the syringe plunger may drop out of the barrel.

60 cm³ of O₂ = 60/24000 = 0.0025 mol

From the second equation, KClO₃ º 1½O₂

So we need 0.0025/1.5 = 0.00167 mol of KClO₃

KClO₃ = 122.5

\ mass of KClO₃ = 122.5 × 0.00167 = 0.20 g

Method

Weigh accurately about 0.20 g of dry potassium chlorate(V) in a weighing boat. Add the potassium chlorate(V) to the side arm test tube. Reweigh the weighing boat to allow the mass transferred to be calculated. Stopper the test tube securely and heat strongly until decomposition is complete. Allow the apparatus to cool completely before recording the volume of oxygen gas collected.

Safety

Potassium chlorate(V) is an oxidizing agent. It must not be contaminated with any other materials as a violent reaction may result. Eye protection should be worn and any spills washed off immediately with water. The glassware will remain hot for some time - care should be taken.

Analysis and evaluation exercise

The experiment was carried out using 0.188 grams of potassium chlorate(V). The volume of gas collected was originally 72 cm³, but on cooling 56.5 cm³ was collected. Work out which is the correct equation for the reaction. Carry out an error analysis on the experiment and suggest possible improvements.

Answer