A 0.8612-g sample of a mixture of NaBr, NaI, and NaNO3 was analyzed by adding AgNO3 and precipitating a 1.0186-g mixture of AgBrand AgI. The precipitate was then heated in a stream of Cl2, convertingit to 0.7125 g of AgCl. Calculate the %w/w NaNO3 in the sample.

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Answer 1

Percentage by weight of the NaNO₃, is given by the ratio of the mass of NaNO₃ to the mass of the sample

The percentage by weight of NaNO₃ in the sample is approximately 30.713 %

Reason:

Given parameters are;

Mass of sample of NaBr, NaI, and NaNO₃ mixture = 0.8612 g

Mass of AgNO₃ and AgI formed = 1.0186

Mass of AgCl formed = 0.7125

Required:

Percentage by weight of NaNO₃ in the sample

Solution:

Let A represent NaBr, let B represent NaI, and let C represent NaNO₃, we have;

Equation 1 A + B + C = 0.8612

The addition of AgNO₃, precipitates 1.0186 g mixture of AgBr and AgI, therefore, we have;

[tex]A \cdot\left(\dfrac{Molar \ mass \ of \ AgBr}{Molar \ mass \ of \ NaBr} \right) + B \cdot \left(\dfrac{Molar \ mass \ of \ AgI}{Molar \ mass \ of \ NaI} \right) = 1.0186[/tex]

Molar mass of AgBr = 187.77 g/mol

Molar mass of NaBr = 102.894 g/mol

Molar mass of AgI = 234.77 g/mol

Molar mass of NaI = 149.89 g/mol

Which gives;

Equation 2 [tex]A \cdot\left(\dfrac{187.77}{102.894} \right) + B \cdot \left(\dfrac{234.77}{149.89} \right) = 1.0186[/tex]

The forming of the precipitate by passing the compounds through a stream of chlorine gives;

[tex]A \cdot\left(\dfrac{Molar \ mass \ of \ AgCl}{Molar \ mass \ of \ NaBr} \right) + B \cdot \left(\dfrac{Molar \ mass \ of \ AgCl}{Molar \ mass \ of \ NaI} \right) = 0.7125[/tex]

Molar mass of AgCl = 143.32 g/mol

Therefore, we get;

Equation 3 [tex]A \cdot\left(\dfrac{143.32}{102.894} \right) + B \cdot \left(\dfrac{143.32}{149.89} \right) = 0.7125[/tex]

Solving equation (2), and (3) gives;

A ≈ 0.3252 g, B ≈ 0.2715 g

Plugging in the values of A, and B in equation (1), gives;

0.3252 + 0.2715 + C = 0.8612

C = 0.8612 - (0.3252 + 0.2715) ≈ 0.2645

The mass of NaNO₃ in the mixture, C ≈ 0.2645

[tex]The \ percent \ by \ weight \ of \ NaNO_3 = \dfrac{0.2645}{0.8612} \times 100 \approx \underline{30.713 \%}[/tex]

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Answer: hope this helps mark me BRAINLEST plz ;)

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