Question

How many moles of nitrite ion in $6.31 \times {10^{ - 1}}{\text{ moles}}$ of ${\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}$?

Answer 1

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

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

Thus, moles of nitrite ion in $6.31 \times {10^{ - 1}}{\text{ moles}}$ of ${\text{Ni}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}$ are $12.62 \times {10^{ - 1}}{\text{ mol}}$.

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.