# Question d34c1

Apr 13, 2015

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, $A g N {O}_{3}$.

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

$A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} \to A g C {l}_{\left(s\right)}$

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

${n}_{C {l}^{-}} = {n}_{A g C l} = \text{0.01744 moles }$ $C {l}^{-}$

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

${\overbrace{{n}_{N a C l}}}^{\textrm{x}} + {\underbrace{{n}_{K C l}}}_{\textrm{y}} = 0.01744 \text{ }$ $\textcolor{b l u e}{\left(1\right)}$

One mole of $N a C l$ produces 1 mole of $C {l}^{-}$ ions; the same is true for $K C l$.

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

${M}_{\text{M NaCl}}$, ${M}_{\text{M KCl}}$ - the molar masses of sodium chloride and potassium chloride, respectively.

$x \cdot 58.44 + y \cdot 74.55 = \text{0.8870 g }$ $\textcolor{b l u e}{\left(2\right)}$,

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

$x = 0.01744 - y$

$58.44 \cdot \left(0.01744 - y\right) + 74.55 \cdot y = 0.8870$

$1.019 - 58.44 \cdot y + 74.55 \cdot y = 0.8870$

$16.11 \cdot y = - 0.132 \implies y = \textcolor{red}{- 0.008194}$

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