# Question ffe51

Sep 10, 2017

$\text{0.24 moles}$

#### Explanation:

For starters, you know that $1$ mole of sodium chloride contains $1$ mole of sodium cations, ${\text{Na}}^{+}$, and $1$ mole of chloride anions, ${\text{Cl}}^{-}$.

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. 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

$0.2396 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaCl"))) * "1 mole Cl"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("0.24 moles Cl}}}}$

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 $\text{14 g}$ of sodium chloride will contain $0.24$ moles of chloride anions, ${\text{Cl}}^{-}$, i.e. $0.24$ moles of chlorine atoms, $\text{Cl}$, reacted to make this sample.