# Question d61b3

May 8, 2017

The sample is 82 % pure.

#### Explanation:

Step 1. Start with the balanced equation.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 58.44 \textcolor{w h i t e}{m m m m m m m m m m m l l} 143.32$
$\textcolor{w h i t e}{m m m} \text{NaCl" + "AgNO"_3 → "NaNO"_3 + "AgCl}$

Step 2. Calculate the moles of $\text{AgCl}$.

$\text{Moles of AgCl" = 2.4 color(red)(cancel(color(black)("g AgCl"))) × ("1 mol AgCl")/(143.32 color(red)(cancel(color(black)("g AgCl")))) = "0.0167 mol AgCl}$

Step 3. Calculate the moles of $\text{NaCl}$

$\text{Moles of NaCl" = 0.0167 color(red)(cancel(color(black)("mol AgCl"))) × ("1 mol NaCl")/(1 color(red)(cancel(color(black)("mol AgCl")))) = "0.0167 mol NaCl}$

Step 4. Calculate the mass of $\text{NaCl}$

$\text{Mass of NaCl" = 0.0167 color(red)(cancel(color(black)("mol NaCl"))) × "58.44 g NaCl"/(1 color(red)(cancel(color(black)("mol NaCl")))) = "0.979 g NaBr}$

Step 5. Calculate the purity of the $\text{NaCl}$

" % Purity" = "Mass of NaCl"/"Total mass" × 100 % = (0.979 color(red)(cancel(color(black)("g"))))/(1.2 color(red)(cancel(color(black)("g")))) × 100 % = 82 %#