What was the percentage of NaCl in the mixture?

Question

A mixture if NaCl and KC1 weighed 5.4892 g. The sample was dissolved in water and reacted with an excess of silver nitrate in solution. The resulting AgCl weighed 12.705 2 g. What was the percentage of NaCl in the mixture?

Answer 1

In the mixture we have Na, K and Cl atoms. Let the mixture be represented by x(NaCl)+y(KCl). The only source of Cl atoms in the silver chloride product is the amount of Cl atoms in the mixture=(x+y)Cl. The amount in grams of Cl in the AgCl is in the same ratio as the atomic weight of Cl to the molecular weight of AgCl. The atomic weight of Cl=35.453, Ag=107.8682, and the molecular weight of AgCl is 107.8682+35.453=143.321 approx. So (35.453/143.321)×12.7052=3.1429g approx. Therefore (x+y)Cl=35.453(x+y) is equivalent to 3.1429g. We can relate this molecular weight (x+y Cl atoms) to the actual weight 3.1429 using the constant n=35.453(x+y)/3.1429. The same constant applies to all the weights.

3.1429g of the mixture is Cl. What remains is made up of Na (atomic weight 22.9893) and K (atomic weight 39.0983) atoms only=5.4892-3.1429=2.3463g approx. So xNa+yK=22.9893x+39.0983y is equivalent to 2.3463 grams, and n=(22.9893x+39.0983y)/2.3463. 

So n=35.453(x+y)/3.1429=(22.9893x+39.0983y)/2.3463

From this equation we can relate x and y:

Cross-multiply:

2.3463×35.453(x+y)=3.1429(22.9893x+39.0983y)

83.1834(x+y)=72.2531x+122.8820y, (83.1834-72.2531)x=(122.8820-83.1834)y

10.9303x=39.6987y and x=3.632y approx.

We need x/(x+y)=3.632y/(3.632y+y)=0.7841 or 78.41%.

So NaCl is 78.41% of the mixture.