Question #d61b3

1 Answer
May 8, 2017

Answer:

The sample is 82 % pure.

Explanation:

Step 1. Start with the balanced equation.

#M_text(r):color(white)(m) 58.44color(white)(mmmmmmmmmmmll)143.32#
#color(white)(mmm)"NaCl" + "AgNO"_3 → "NaNO"_3 + "AgCl"#

Step 2. Calculate the moles of #"AgCl"#.

#"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 #"NaCl"#

#"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 #"NaCl"#

#"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 #"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 %#