Question #af337

The balanced chemical equation that describes this synthesis reaction is the key here.

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

Notice that the reaction consumes #color(blue)(2)# moles of sodium for every #1# mole of chlorine gas that takes part in the reaction and produces #color(purple)(2)# moles of sodium chloride.

In your case, you know that #6.7# moles of sodium are allowed to react with #3.2# moles of chlorine gas. In order to figure out which reactant acts as a limiting reagent, you need to see if you have enough moles of one reactant to ensure that all the moles of the second reactant can actually take part in the reaction.

So, pick one reactant and use the mole ratio that exists between the two reactants to see what you can find.

#6.7 color(red)(cancel(color(black)("moles Na"))) * "1 mole Cl"_2/(color(blue)(2)color(red)(cancel(color(black)("moles Na")))) = "3.35 moles Cl"_2#

This means that in order for all the moles of sodium to react, you must provide #3.35# moles of chlorine gas, Since you don't have enough moles of chlorine gas available

#overbrace("3.35 moles Cl"_2)^(color(red)("what you need")) " " > " " overbrace("3.2 moles Cl"_2)^(color(red)("what you have"))#

you can say that chlorine gas will act as the limiting reagent, i.e. it will be completely consumed before all the moles of sodium will get the chance to react.

So, you know that the reaction consumes #3.2# moles of chlorine gas, which implies that it produces

#3.2 color(red)(cancel(color(black)("moles Cl"_2))) * (color(purple)(2)color(white)(.)"moles NaCl")/(1color(red)(cancel(color(black)("mole Cl"_2)))) = color(darkgreen)(ul(color(black)("6.4 moles NaCl")))#