Question #6cd19

1 Answer
Aug 9, 2016

Answer:

#"263 g"#

Explanation:

As you know, one formula unit of sodium chloride, #"NaCl"#, contains

  • one sodium cation, #1 xx "Na"^(+)#
  • one chloride anion, #1 xx "Cl"^(-)#

On the other hand, one formula unit of magnesium chloride, #"MgCl"_2#, contains

  • one magnesium cation, # 1 xx "Mg"^(2+)#
  • two chloride anions, # 2 xx "Cl"^(-)#

This means that for one mole of sodium chloride you're getting #2# moles of ions, and for one mole of magnesium chloride you're getting #3# moles of ions.

In other words, the number of ions present in one mole of magnesium chloride, i.e. #3# moles of ions, is also present in #3/2# moles of sodium chloride, since

#3/2 color(red)(cancel(color(black)("moles NaCl"))) * "2 moles ions"/(1color(red)(cancel(color(black)("mole NaCl")))) = "3 moles ions"#

All you have to do now is calculate how many moles of magnesium chloride you have in your sample

#285 color(red)(cancel(color(black)("g NaCl"))) * "1 mole MgCl"_2/((24 + 2 xx 35.5)color(red)(cancel(color(black)("g NaCl")))) = "3 moles MgCl"_2#

Since you know that you need #3/2# moles of sodium chloride to get the same number of ions as #1# mole of magnesium chloride, you can say that you have

#3 color(red)(cancel(color(black)("moles MgCl"_2))) * (3/2color(white)(a) "moles Na")/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = 9/2color(white)(a) "moles NaCl"#

Finally, use the molar mass of sodium chloride to convert this to grams

#9/2 color(red)(cancel(color(black)("moles NaCl"))) * ((23 + 35.5)color(white)(a)"g")/(1color(red)(cancel(color(black)("mole NaCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("263 g NaCl")color(white)(a/a)|)))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of magnesium chloride.