Question #ffe51

For starters, you know that #1# mole of sodium chloride contains #1# mole of sodium cations, #"Na"^(+)#, and #1# mole of chloride anions, #"Cl"^(-)#.

This is equivalent to saying that in order to have #1# mole of sodium chloride, you need to combine #1# mole of sodium atoms and #1# mole of chlorine atoms.

![www.dvusd.org]()

In order to find the number of moles of sodium chloride present in your sample, you can use the molar mass of this compound, i.e. the mass of exactly #1# mole of sodium chloride.

You will have

#14 color(red)(cancel(color(black)("g"))) * overbrace("1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))))^(color(blue)("the molar mass of NaCl")) = "0.2396 moles NaCl"#

This means that your sample will contain

#0.2396 color(red)(cancel(color(black)("moles NaCl"))) * "1 mole Cl"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("0.24 moles Cl")))#

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

So, you can say that #"14 g"# of sodium chloride will contain #0.24# moles of chloride anions, #"Cl"^(-)#, i.e. #0.24# moles of chlorine atoms, #"Cl"#, reacted to make this sample.