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Read the following passage and answer the questions given below :
Most of metals occur in combined state in form of ores. Carbonate ores are converted into oxides by calcination and sulphide ores by roasting. Oxides are reduced with suitable reducing agent like carbon to get free metal. Highly reactive metals like – $\ce{Al}$, $\ce{Mg}$ are also used as reducing agents to obtain metal from their oxides. Most reactive metals are obtained by electrolytic reduction of their molten ores. Alloying is a very good method of improving the properties of a metal. We can get desired properties by this method. The electrical conductivity and melting point of an alloy is less than that of pure metals.
(a) Why carbonate or sulphide ores are converted to oxides before extraction of metal from it?
(b) Write a reaction in which Aluminium is used as a reducing agent to obtain metal from its oxide.
Most of metals occur in combined state in form of ores. Carbonate ores are converted into oxides by calcination and sulphide ores by roasting. Oxides are reduced with suitable reducing agent like carbon to get free metal. Highly reactive metals like – $\ce{Al}$, $\ce{Mg}$ are also used as reducing agents to obtain metal from their oxides. Most reactive metals are obtained by electrolytic reduction of their molten ores. Alloying is a very good method of improving the properties of a metal. We can get desired properties by this method. The electrical conductivity and melting point of an alloy is less than that of pure metals.
(a) Why carbonate or sulphide ores are converted to oxides before extraction of metal from it?
(b) Write a reaction in which Aluminium is used as a reducing agent to obtain metal from its oxide.
Show SolutionHide Solution↓
(a) Because it is easier to obtain metal from its oxide. /
Because it is easier to reduce metal oxide to metal (1)
(b)
$\ce{Fe2O3(s)+ 2Al(s) -> 2Fe(l) + Al2O3(s) + Heat}$
$\ce{3MnO2(s) + 4Al(s) -> 3Mn(l) + 2Al2O3(s) + Heat}$ (1)
(balancing is optional) (any one equation)
(c) (i) $\ce{2Cu2S + 3O2(g) ->[Heat] 2Cu2O(s) + 2SO2 (g)}$ (1)
$\ce{2Cu2O + Cu2S ->[Heat] 6Cu(s) + SO2(g)}$ (1)
OR
(c)(ii) (I) Because highly reactive metals have more affinity for
oxygen than carbon. (1)
(II) Because of its low melting point. (1)
Because it is easier to reduce metal oxide to metal (1)
(b)
$\ce{Fe2O3(s)+ 2Al(s) -> 2Fe(l) + Al2O3(s) + Heat}$
$\ce{3MnO2(s) + 4Al(s) -> 3Mn(l) + 2Al2O3(s) + Heat}$ (1)
(balancing is optional) (any one equation)
(c) (i) $\ce{2Cu2S + 3O2(g) ->[Heat] 2Cu2O(s) + 2SO2 (g)}$ (1)
$\ce{2Cu2O + Cu2S ->[Heat] 6Cu(s) + SO2(g)}$ (1)
OR
(c)(ii) (I) Because highly reactive metals have more affinity for
oxygen than carbon. (1)
(II) Because of its low melting point. (1)