Question

Question #d34c1

Answer 1

I think the information you were given is incorrect. More specifically, I suspect that either the mass of the sample is too small, or the mass of the formed precipitate is too big.

Here's why.

So, you have a mixture of sodium chloride and potassium chloride which you react, in aqueous solution, with silver nitrate, `AgNO_3`.

The reaction will lead to the formation of silver chloride, a solid that will precipitate out of solution. The net ionic equation for this reaction describes what's going on

`Ag_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s))`

Notice that 1 mole of chloride ions is needed to produce 1 mole of silver chloride precipitate. This means that you can calculate the total number of moles of chloride that reacted by using the mass of the precipitate

`2.5cancel("g") * "1 mole AgCl"/(143.32cancel("g")) = "0.01744 moles AgCl"`

Since you have a `1:1` mole ratio between silver chloride and the chloride ions, you'll have

`n_(Cl^(-)) = n_(AgCl) = "0.01744 moles "` `Cl^(-)`

This represents the total number of moles of chloride present in the sample. Since both sodium chloride and potassium chloride dissociate completely in aqueous solution, you can write

`overbrace(n_(NaCl))^text(x) + underbrace(n_(KCl))_text(y) = 0.01744" "` `color(blue)((1))`

One mole of `NaCl` produces 1 mole of `Cl^(-)` ions; the same is true for `KCl`.

The second equation will use the mass of the sample, which can be written like this

`x * M_("M NaCl") + y * M_("M KCl") = "0.8870 g "`, where

`M_("M NaCl")`, `M_("M KCl")` - the molar masses of sodium chloride and potassium chloride, respectively.

`x * 58.44 + y * 74.55 = "0.8870 g "` `color(blue)((2))`,

So far so good, but here is where the problem stops making sense. Trying to solve the two equations will lead to

`x = 0.01744 - y`

`58.44 * (0.01744 - y) + 74.55 * y = 0.8870`

`1.019 - 58.44 * y + 74.55 * y = 0.8870`

`16.11 * y = -0.132 => y = color(red)(-0.008194)`

Since `y` represents number of moles, it cannot be negative. This is the problem with the data given.

The only way to get positive values for both `x` and `y` is to either

• reduce the mass of silver chloride, which will reduce the number of moles of chloride present in solution
• increase the mass of the sample