Question

(a) The diagram given below is to prepare iron (III) chloride in the laboratory:
Figure

(i) What is substance B?
(ii) What is the purpose of B?
(iii) Why is iron (III) chloride to be stored in a closed container?
(iv) Write the equation for the reaction between iron and chloride.
(b) Write the equation(s) for the reaction(s) to prepare lead sulphate from lead carbonate.

Answer 1

(i) Substance B is anhydrous (fused) calcium chloride (${\mathrm{CaCl}}_2$).

(ii) Anhydrous calcium chloride absorbs the moisture present in the apparatus. This helps in keeping the anhydrous ferric chloride dry as it is a highly deliquescent salt.

(iii) Iron (III) chloride should be stored in a closed container because of its deliquescent nature. In the presence of moisture, iron (III) chloride absorbs water, gets dissolved in it and forms a saturated solution.

(iv) When dried chlorine gas is passed over heated iron, it forms anhydrous iron (III) chloride.

$$\begin{gathered} 2\mathrm{Fe}+3{\mathrm{Cl}}_2\to 2{\mathrm{FeCl}}_3 \\ \mathrm{Anhydrous} \\ \mathrm{iron}\left(\mathrm{III}\right)\mathrm{chloride} \end{gathered}$$

(b) The preparation of lead sulphate from lead carbonate involves the following steps:
Step I: Insoluble lead carbonate first reacts with dilute nitric acid to form soluble lead nitrate.

$$\begin{gathered} {\mathrm{PbCO}}_3+\mathrm{dil}.2{\mathrm{HNO}}_3\stackrel{}{\to }\mathrm{Pb}({\mathrm{NO}}_3{)}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2 \\ \mathrm{Lead}\mathrm{carbonate}\mathrm{Nitric}\mathrm{acid}\mathrm{Lead}\mathrm{nitrate}(\mathrm{Soluble}) \end{gathered}$$

​ Step II: Dilute sulphuric acid is then added to lead nitrate to form lead sulphate and nitric acid.

$$\begin{gathered} \mathrm{Pb}({\mathrm{NO}}_3{)}_2+\mathrm{dil}.{\mathrm{H}}_2{\mathrm{SO}}_4\stackrel{}{\to }{\mathrm{PbSO}}_4+2{\mathrm{HNO}}_3 \\ \mathrm{Lead}\mathrm{nitrate}\mathrm{Sulphuric}\mathrm{acid}\mathrm{Lead}\mathrm{sulphate}\mathrm{Nitric}\mathrm{acid} \end{gathered}$$