Question

a green inorganic compound in the form of crystals when heated in a test tube produces reddish brown Residue with characteristics smell of burning sulphur (i) identify compound
(ii) give balanced chemical equation.
(iii) identify type of equation

Answer 1

Procedure:

Real Lab Procedure:

Take about 2g of ferrous sulphate crystals in a dry boiling tube and note the colour of the crystals.Hold the boiling tube with a test tube holder and heat the boiling tube over the flame of a burner.Smell the gas being emitted.Observe the colour of the crystals after heating.

To use the simulator:

You can select the type of reaction form ‘Select the type of reaction’ drop down list.

Thermal Decomposition

You can select the sample form ‘Select the sample’ drop down list.To start the experiment, click on the ‘Start’ button.To put copper carbonate into the test tube, drag the spatula over the test tube.To turn on the burner.

Answer 2

Answer:

For (i): the compound is iron sulfate.

For (ii): The chemical equation is given below.

For (iii): The equation is a type of decomposition reaction.

Explanation:

• For (i):

When a green colored inorganic compound is heated, it produces reddish brown residue and releases a gas having a small of sulfur.

The green colored compound is iron sulfate and reddish brown residue is ferric oxide.

• For (ii):

The chemical equation for the heating of ferrous sulfate is given by the equation:

$2FeSO_4\overset{\Delta}\rightarrow Fe_2O_3+SO_2+SO_3$

By Stoichiometry of the reaction:

2 moles of iron sulfate produces 1 mole of ferric oxide, 1 mole of sulfur dioxide and 1 mole of sulfur trioxide.

• For (iii):

This reaction is a type of thermal decomposition reaction because one single substance is breaking don into two or more smaller substances in the presence of heat.