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Question: Find out the mole ratio of \({{C}_{2}}{{O}_{4}}^{2-}\)in the compound. A.\(1.13\times {{10}^{-3}}\...

Find out the mole ratio of C2O42−{{C}_{2}}{{O}_{4}}^{2-}in the compound.
A.1.13×10−31.13\times {{10}^{-3}}
B.1.13×10−21.13\times {{10}^{-2}}
C.2.26×10−32.26\times {{10}^{-3}}
D.2.26×10−22.26\times {{10}^{-2}}

Explanation

Solution

Before answering this question, we should know that Whenever reactants react in a chemical reaction, they have a set ratio. The ratio should be balanced else there will be the leftover reactant. For better knowledge of this, we should know about the mole ratio.

Complete answer:
The ratio of amounts in moles of any two compounds that are reacting in the chemical reaction is the mole ratio. It can be obtained by looking at the coefficients in a balanced chemical equation. It is also known as the mole-to-mole ratio. They are also considered as conversion factors of products and reactants.
When we study the relative amounts of substances in a chemical reaction or stoichiometry, there will be two circumstances where we will need to calculate the mole ratio. When we are finding an unknown substance to find its empirical formula and the other situation is when we are calculating relative amounts of reactants and products in a chemical reaction.
Let’s consider a solution having a mol and b mol of Cu2+C{{u}^{2+}} and C2O42−{{C}_{2}}{{O}_{4}}^{2-} ions.
Only C2O42−{{C}_{2}}{{O}_{4}}^{2-}gets oxidized by KMnO4KMn{{O}_{4}} in the reaction.
The reaction that is gonna take place here is Redox :
Mn7++5e−→Mn2+M{{n}^{7+}}+5{{e}^{-}}\to M{{n}^{2+}}
C23+→2C4++2eC_{2}^{3+}\to 2{{C}^{4+}}+2e
Meq of C2O42−= Meq of KMnO4{{M}_{eq}}\,of\,{{C}_{2}}{{O}_{4}}^{2-}=\,{{M}_{eq}}\,of\,KMn{{O}_{4}}
b×2×1000=0.02×5×22.6b\times 2\times 1000=0.02\times 5\times 22.6
b= 1.13×10−31.13\times {{10}^{-3}}

So, Option (A) 1.13×10−31.13\times {{10}^{-3}} is correct.

Note:
Let’s take one example :
In reaction:
2H2(g)+O2(g)→2H2O(g)2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(g)
In this 2:1 is the mole ratio between H2O and O2{{H}_{2}}O\,and\,{{O}_{2}}. Every 1 mole of O2{{O}_{2}}used, there will be 2 moles of H2O{{H}_{2}}Othat will be formed.
And 1:1 is the mole ratio between H2O and H2{{H}_{2}}O\,and\,{{H}_{2}}. Every 2 moles of H2{{H}_{2}}used, there will be 2 moles of H2O{{H}_{2}}Othat will be formed. If 4 moles of hydrogen is taken, then 4 moles of water will be formed.