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A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 1.903 g of AgCl. Calculate the percent by mass of each compound in the mixture.

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Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess
AgNO_3 then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.


\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=(xg)/(58.5g/mole)=(x)/(58.5)moles


\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=((0.9440-x)g)/(74.5g/mole)=((0.9440-x))/(74.5)moles

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl =
(x)/(58.5)moles

Moles of chloride ions in KCl =
((0.9440-x))/(74.5)moles

The total moles of chloride ions =
(x)/(58.5)moles+((0.9440-x))/(74.5)moles

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion =
(x)/(58.5)moles+((0.9440-x))/(74.5)moles

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole


\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=(1.903g)/(143.32g/mole)=0.0133moles

Now we have to determine the value of 'x'.

Moles of AgCl =
(x)/(58.5)moles+((0.9440-x))/(74.5)moles

0.0133 mole =
(x)/(58.5)moles+((0.9440-x))/(74.5)moles

By solving the term, we get the value of 'x'.


x=0.171g

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.


\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}* 100=(0.171g)/(0.9440g)* 100=18.11\%


\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}* 100=(0.773g)/(0.9440g)* 100=81.88\%

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

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