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Question: How many moles of nitrite ion in \(6.31 \times {10^{ - 1}}{\text{ moles}}\) of \({\text{Ni}}{\left( ...

How many moles of nitrite ion in 6.31×10−1 moles6.31 \times {10^{ - 1}}{\text{ moles}} of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}?

Explanation

Solution

To solve this first write the balanced chemical reaction for the dissociation of the given compound i.e. Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}. Then from the reaction stoichiometry calculate the mole ratio between Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} and the nitrite ion. From the mole ratio, calculate the number of moles of nitrite ion in 6.31×10−1 moles6.31 \times {10^{ - 1}}{\text{ moles}} of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}.

Complete solution:
To solve this first write the balanced chemical reaction for the dissociation of the given compound i.e. Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}. Then from the reaction stoichiometry calculate the mole ratio between Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} and the nitrite ion. From the mole ratio, calculate the number of moles of nitrite ion in 6.31×10−1 moles6.31 \times {10^{ - 1}}{\text{ moles}} of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}.
The dissociation reaction of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} is as follows:
Ni(NO2)2→Ni2++2NO2−{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} \to {\text{N}}{{\text{i}}^{2 + }} + {\text{2NO}}_2^ -
From the balanced reaction stoichiometry, we can say that one mole of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} corresponds to two moles of nitrite ions i.e. NO2−{\text{NO}}_2^ - ions. Thus,
1 mol Ni(NO2)2=2 mol NO2−{\text{1 mol Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} = 2{\text{ mol NO}}_2^ -
Now calculate the number of moles of nitrite ions i.e. NO2−{\text{NO}}_2^ - ions in 6.31×10−1 moles6.31 \times {10^{ - 1}}{\text{ moles}} of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} as follows:
As stated above, from the balanced reaction stoichiometry, we can say that one mole of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} corresponds to two moles of nitrite ions i.e. NO2−{\text{NO}}_2^ - ions. Thus,
1 mol Ni(NO2)2=2 mol NO2−{\text{1 mol Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} = 2{\text{ mol NO}}_2^ -
Thus, the number of moles of nitrite ions i.e. NO2−{\text{NO}}_2^ - ions in 6.31×10−1 moles6.31 \times {10^{ - 1}}{\text{ moles}} of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} are,
Number of moles of nitrite ions =6.31×10−1 mol Ni(NO2)2×2 mol NO2− 1 mol Ni(NO2)2 = 6.31 \times {10^{ - 1}}{\text{ mol Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} \times \dfrac{{2{\text{ mol NO}}_2^ - }}{{{\text{ 1 mol Ni}}{{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)}_{\text{2}}}}}
Number of moles of nitrite ions =12.62×10−1 mol = 12.62 \times {10^{ - 1}}{\text{ mol}}

Thus, moles of nitrite ion in 6.31×10−1 moles6.31 \times {10^{ - 1}}{\text{ moles}} of Ni(NO2)2{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}} are 12.62×10−1 mol12.62 \times {10^{ - 1}}{\text{ mol}}.

Note: Remember that the balanced chemical equation for the given reaction must be written correctly. Incorrect or unbalanced chemical equations can lead to incorrect number of moles which can lead to incorrect mass of the element.