Question #92995

1 Answer
Mar 15, 2016

#"322 g Na"#


Your starting point here will be the balanced chemical equation for this synthesis reaction

#color(blue)(2)"Na"_text((s]) + "Cl"_text(2(g]) -> color(red)(2)"NaCl"_text((s])#

The interesting thing to notice here is that you can determine how much sodium reacted simply by using the mass of sodium chloride produced by the reaction.

If you assume that the reaction has a #100%# yield, which is a reasonable assumption given the information provided by the problem, you can always find how much sodium reacted by looking at how much sodium chloride was produced.

So, #color(blue)(2)# moles of sodium metal will require #1# mole of chlorine gas to produce #color(red)(2)# moles of sodium chloride. In other words, for every two moles of sodium chloride produced by the reaction, two moles of sodium metal and one mole of chlorine gas were consumed.

You know how much sodium chloride was produced, so use the compound's molar mass to convert this to moles.

#819color(red)(cancel(color(black)("g"))) * overbrace("1 mole NaCl"/(58.443color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of NaCl")) = "14.014 moles NaCl"#

Now, use the number of moles of sodium chloride to determine how much sodium metal reacted. Since you have a #color(blue)(2):color(red)(2)# mole ratio between sodium metal and sodium chloride, it follows that the reaction must have consumed

#14.014color(red)(cancel(color(black)("moles NaCl"))) * (color(blue)(2)color(white)(a)"moles Na")/(color(red)(2)color(red)(cancel(color(black)("moles NaCl")))) = "14.014 moles Na"#

To determine how many grams of sodium metal would contain this many moles, use the element's molar mass

#14.014color(red)(cancel(color(black)("moles Na"))) * "22.99 g"/(1color(red)(cancel(color(black)("mole Na")))) = color(green)(|bar(ul(color(white)(a/a)"322 g"color(white)(a/a)|)))#

The answer is rounded to three sig figs.

So, how much chlorine gas did the reaction consume?

Once again, use the mole ratio that exists between sodium chloride and chlorine gas

#14.014 color(red)(cancel(color(black)("moles NaCl"))) * "1 mole Cl"_2/(color(red)(2)color(red)(cancel(color(black)("moles NaCl")))) = "7.007 moles Cl"_2#

The mass of chlorine gas that contains this many moles is

#7.007 color(red)(cancel(color(black)("moles Cl"_2))) * "70.906 g"/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "496.8 g Cl"_2#

As you can see, the reaction consumed almost all the moles of chlorine gas.