# Question 9d84d

Aug 14, 2017

$\text{5 g NaCl}$

#### Explanation:

Start by writing the balanced chemical equation that describes this reaction

$2 {\text{Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl}}_{\left(s\right)}$

Notice that the reaction consumes $2$ moles of sodium metal for every $1$ mole of chlorine gas that takes part in the reaction and produces $2$ moles of sodium chloride.

Use the molar masses of the two reactants to convert the samples to moles

22 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23.0color(red)(cancel(color(black)("g")))) = "0.9565 moles Na"

3 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "0.04231 moles Cl"_2

At this point, it should be clear that chlorine gas will act as a limiting reagent because you don't have nearly enough moles chlorine gas to ensure that all the moles of sodium can react.

More specifically, that much sodium would require

0.9565 color(red)(cancel(color(black)("mole Na"))) * "1 mole Cl"_2/(2color(red)(cancel(color(black)("moles Na")))) = "0.4783 moles Cl"_2

Since

overbrace("0.04231 moles Cl"_2)^(color(blue)("what you have")) " " < " " overbrace("0.4783 moles Cl"_2)^(color(blue)("what you need"))

you can say that the reaction will consume all the moles of chlorine gas and only

0.04231 color(red)(cancel(color(black)("moles Cl"_2))) * "2 moles Na"/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "0.08462 moles Na"

and produce

0.04231 color(red)(cancel(color(black)("moles Cl"_2))) * "2 moles NaCl"/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "0.08462 moles NaCl"#

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

$0.08462 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("5 g}}}}$

The answer is rounded to one significant figure, the number of sig figs you have for the mass of chlorine gas.