Question

A mixture of $NaI$ and $NaCl$ gave ${H_2}S{O_4}$ , produced sodium sulfate equal to the weight of the original mixture taken. Find % of $NaI$ in the mixture.

Answer 1

To solve this question, we need to form two equations based on the question where first we will consider mass of mixture before heat was applied and the other when heat was applied. Then, we will subtract both the equations to find the values of X and Y by substituting the atomic masses. Finally, find the percentage of the mixture.

Complete step by step answer:

Here, we will consider the mass of mixture =$100gm$
So, it will produce $100gm$ of $N{a_2}S{O_4}$
Therefore, the mass of $Na$ in one molecule = $\dfrac{{mass of Na}}{{molar mass}}$ = $\dfrac{{2 \times 23}}{{142}}g/mol = 0.34g/mol$
Now, the mass of $Na$ will become = $100 \times 0.34 = 32.4gm$
Before heating the mass of I and Cl will be equal to the mass of sulfate molecule
So, mass of sulfate =$100 - 32.4 = 67.6g$
We know that, at the beginning mass of I and Cl was $67.6g.$
Now, contribution of I and $Cl$ in $NaI$ and $NaCl$ is calculated by considering the contribution of I will be ‘X’ mole and for $Cl$ is ‘Y’ moles
then, the equation becomes $X \times I + Y \times Cl = 67.6$
Now, we will put the atomic masses of I and $Cl$ i.e.$127\;{\text{and 35}}{\text{.5}}$ respectively,
Then, equation (1) becomes $X \times 127 + Y \times 35.5 = 67.6$
Likewise, the equation for contribution of sodium will be
$X \times Na + Y \times Na = 32.4$
Eq(2)$X \times 23 + Y \times 23 = 32.4$ Since atomic mass of $Na$=$23$
Now, we will subtract both the equation eq(2) – eq(1):
$127x{\text{ }} + {\text{ }}35.5y{\text{ }} = {\text{ }}67.6$
$23x{\text{ }} - {\text{ }}23{\text{ }}y{\text{ }} = {\text{ }} - 34.2{\text{ }}]{\text{ }}x{\text{ }}1.54$

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$\;91.5x{\text{ }} = {\text{ }}15$ $$$$
Therefore, value of X = $0.16$
Then, value of Y =$\dfrac{{[\left( {32.4{\text{ }} - {\text{ }}23{\text{ }}} \right)x{\text{ }}0.16]}}{{23}}$ =$1.24mol$
Now, the molar mass of $NaI$ will become $= {\text{ }}23{\text{ }} + 123{\text{ }} = {\text{ }}150{\text{ }}g$
As it contains $0.16mol$ of $NaI$ then the new mass will become $= {\text{ }}150{\text{ }}x{\text{ }}0.16{\text{ }} = {\text{ }}24{\text{ }}g$
Therefore,$100g$sample of mixture will contain $24g$ of $NaI$
Thus, percentage of $NaI$= $24\%$

Note:
In the Finkelstein reaction, sodium iodide is used for the conversion of alkyl chlorides into alkyl iodides. The reaction relies on the insolubility of sodium chloride in acetone to complete the reaction. The equation of Finkelstein reaction is:
$R - Cl + NaI \to R - I + NaCl$