Question #38980

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Question #38980

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Answer 1 · 2016-03-13T15:21:03

${CoSO}_{4} + {Pb(NO}_{3})_{2} \to {Co(NO}_{3})_{2} + {PbSO}_{4}$ $"CoSO"_4+"Pb(NO"_3)_2->"Co(NO"_3)_2+"PbSO"_4$

Explanation:

Reactants:
1. Cobalt (II) $\Rightarrow$ $=>$ An ion of Cobalt, Co
having valency of $2 + \Rightarrow {Co}^{2 +}$ $2+ =>"Co"^(2+)$ and Sulfate $\Rightarrow {SO}_{4}^{2 -}$ $=>"SO"_4^(2-)$ ion
Combining both the chemical formula of first reactant is ${CoSO}_{4}$ $"CoSO"_4$
2. Similarly Lead (II) $\Rightarrow {Pb}^{2 +}$ $=>"Pb"^(2+)$
and Nitrate $\Rightarrow {NO}_{3}^{-}$ $=>"NO"_3^-$ ion
Observe that each ion of Lead will require two ions of nitrate to satisfy (quench) its valency. Therefore, combining both we obtain chemical formula of second reactant as ${Pb(NO}_{3})_{2}$ $"Pb(NO"_3)_2$

Products:
Following the same logic as used for reactants
1. Cobalt (II) Nitrate has chemical formula as ${Co(NO}_{3})_{2}$ $"Co(NO"_3)_2$
and
2. Lead (II) Sulfate has chemical formula as ${PbSO}_{4}$ $"PbSO"_4$

Writing the chemical equation of the reaction
Reactants $\to$ $->$ Products

${CoSO}_{4} + {Pb(NO}_{3})_{2} \to {Co(NO}_{3})_{2} + {PbSO}_{4}$ $"CoSO"_4+"Pb(NO"_3)_2->"Co(NO"_3)_2+"PbSO"_4$
Inspection reveals that the equation is already balanced.

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Question #38980

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