Question #9d84d

Start by writing the balanced chemical equation that describes this reaction

#2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))#

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 color(red)(cancel(color(black)("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.