4.130 a 0.8870-G Sample of a Mixture of Nacl and Kcl Is Dissolved in Water, and the Solution

4.130 A 0.8870-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 1.913 g of AgCl. Calculate the percent by mass of each compound in the mixture.

4.130 A 0.8870-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 1.913 g of AgCl. Calculate the percent by mass of each compound in the mixture.

First we need a balanced equation. We can just write a general equation, since both of the salts have 1:1 cation to anion ratios.

XCl + AgNO3 à AgCl + XNO3

We know how many moles of AgCl are produced, so from that we can get the number of moles of XCl (right now we’re not worried about what the X is exactly).

Now you know how many moles of XCl there were in the beginning. We also know that this number represents a mixture, so if we call one part of the mixture “x”, then the other part will be the total – “x”.

Let’s let NaCl be “x”

molesNaCl + molesKCl = 0.01334 mol à “x” + molesKCl = 0.01334 mol

molesKCl = 0.01334 mol –x

We can also easily set up a mass relationship now:

massNaCl + massKCl = 0.8870 g

Combining these two, we get something that looks kind of like this:

X = 6.673x10-3 = moles NaCl

SO: mol KCl = 0.01334 – x = 0.01334 - 6.673x10-3 = 6.66710-3 mol KCl

Since the question asks us to find the percent by mass, we should find mass, then percent.

Now we’re just left with percentages, which we can find by taking the part over the whole (the total mass of our sample).