Question

In the reaction, ${{\text{P}}_4} + {\text{NaOH}} \to {\text{P}}{{\text{H}}_3} + {\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2}$ , mole ratio of ${\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2}$ and ${\text{P}}{{\text{H}}_3}$ is:

Answer 1

In order to find mole ratio, we need to first balance the chemical equation to find exact number of moles of ${\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2}$ and ${\text{P}}{{\text{H}}_3}$ produced in the reaction. Then we shall use the stoichiometric coefficients to find the ratio.

Complete step by step answer:
The substance that undergoes a chemical change in the reaction is known as the reactants and new substance formed during the reaction is known as product. A chemical equation represents a chemical reaction.
To balance the chemical equation by oxidation number method:
-Write the skeleton equation representing the chemical change.
-Assign the oxidation number to find out which atoms are undergoing oxidation and reduction, write a separate equation for the atoms undergoing oxidation and reduction.
-Find the change in oxidation number in each equation. Make the change equal in both the equation by multiplying with suitable integers. Add both the equations.
-Now balance those elements which are not undergoing oxidation or reduction followed by balancing of hydrogen and oxygen with the help of water.
-To balance the given chemical reaction in the question:
${{\text{P}}_4} + {\text{NaOH}} \to {\text{P}}{{\text{H}}_3} + {\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2}$
${{\text{P}}_4}\left( 0 \right) + {\text{NaOH}} \to {\text{P}}{{\text{H}}_3}\left( { - 3} \right) + {\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2}\left( { + 1} \right)$ , as we can see it is a disproportionate reaction, in this phosphorus is oxidize to $+ 1$ state and reduces to $- 3$ state. Oxidation half of reaction can be written as: ${{\text{P}}_4} \to {\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2}$ and reduction half reaction can be written as: ${{\text{P}}_4} \to {\text{P}}{{\text{H}}_3}$ .
In order to make the change in oxidation number the same, we can multiply the oxidation half reaction by 3. Now adding both the reduction and oxidation half, we get: $3{{\text{P}}_4} + {{\text{P}}_4} \to 3{\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2} + {\text{P}}{{\text{H}}_3}$ .
Now balancing the element phosphorus and sodium followed by balancing of hydrogen and oxygen. Balanced chemical reaction will be: ${{\text{P}}_4} + 3{\text{NaOH}} + 3{{\text{H}}_2}{\text{O}} \to 3{\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2} + {\text{P}}{{\text{H}}_3}$ .

Hence, mole ratio of ${\text{Na}}{{\text{H}}_2}{\text{P}}{{\text{O}}_2}$ and ${\text{P}}{{\text{H}}_3}$ is equal to $\dfrac{3}{1}$ or 3.

Note:
The total mass of the elements of products of chemical reaction has to be equal to the total mass of the elements of the reactants. So the number of atoms of each element remains the same before and after the reaction.