Using the following unbalanced equation, answer the following questions? #"CaCO"_3(s) + "C"("gr") stackrel(Delta" ")(->) "CaC"_2(s) + "CO"_2(g)#

I will intentionally use a method that makes you think about extensive and intensive properties...

The unbalanced reaction was:

#"CaCO"_3(s) + "C"("gr") stackrel(Delta" ")(->) "CaC"_2(s) + "CO"_2(g)#

#a)# Balance the reaction. I chose calcium first:

#color(red)(2)"CaCO"_3(s) + "C"("gr") stackrel(Delta" ")(->) color(red)(2)"CaC"_2(s) + "CO"_2(g)#

Then I chose oxygen:

#color(red)(2)"CaCO"_3(s) + "C"("gr") stackrel(Delta" ")(->) color(red)(2)"CaC"_2(s) + color(red)(3)"CO"_2(g)#

Then I chose carbon at the end since #"C"# is easy to balance when a single atom is on one side to use.

#color(blue)(color(red)(2)"CaCO"_3(s) + color(red)(5)"C"("gr") stackrel(Delta" ")(->) color(red)(2)"CaC"_2(s) + color(red)(3)"CO"_2(g))#

#b)# Making #"5.0 mols"# #"CO"_2# requires #bb(5/3)# times as many mols of #"CO"_2# as was stated in the reaction (#3#).

Therefore, since #"3 mols"# #"CO"_2# requires #"5 mols C"#,

#5/3 xx "3 mols CO"_2 = "5 mols CO"_2# requires

#5/3 xx "5 mols C" = color(blue)(25/3 "mols")# #"C"# atom.

#c)# We are told that #"CaCO"_3# is in excess, so clearly, #"C"# is the limiting reactant. Therefore, since #"5 mols C"# form #"2 mols CaC"_2#, we scale the reaction down to get:

#4/5 xx (color(red)(2)"CaCO"_3(s) + color(red)(5)"C"("gr") stackrel(Delta" ")(->) color(red)(2)"CaC"_2(s) + color(red)(3)"CO"_2(g))#

Now we read directly from the re-scaled balanced reaction that:

#=> 4/5 xx "5 mols C"# forms #4/5 xx 2 = color(blue)(8/5 "mols CaC"_2)#.

#d)# Here we are told that #"C"# is in excess, so #"CaCO"_3# is the limiting reactant. We have:

#55 cancel("g CaCO"_3) xx ("1 mol CaCO"_3)/(40.08 + 12.011 + 3 cdot15.999 cancel("g CaCO"_3)) = "0.5495 mols CaCO"_3#

Since we have a bit more than #1/2# of a #"mol"# of #"CaCO"_3#, we should make a bit more than #3/4"mols"# of #"CO"_2#:

#0.5495/2 xx (color(red)(2)"CaCO"_3(s) + color(red)(5)"C"("gr") stackrel(Delta" ")(->) color(red)(2)"CaC"_2(s) + color(red)(3)"CO"_2(g))#

Now we read directly from the re-scaled balanced reaction that:

#=> 0.5495/2 xx "2 mols CaCO"_3# makes #0.5495/2 xx "3 mols CO"_2#,

or #color(blue)("0.82 mols CO"_2)# to two sig figs.

As we said, it is indeed a bit more than #3/4 = 0.75# #"mols"# of #"CO"_2#.

#e)# Since #"55 g CaCO"_3# was #"0.5495 mols"#, #"453 g"# of it is

#453/55 xx "0.5495 mols" = "4.526 mols CaCO"_3#

From the mol ratio, we then get that #"5 mols C":"2 mols CaCO"_3# gives:

#("5 mols C")/("2 mols CaCO"_3) = ("? mols C")/("4.526 mols CaCO"_3)#

#=> "11.31 mols C"#

And this has a mass somewhat under #12^2 = "144 g"#:

#11.31 cancel"mols C" xx ("12.011 g C")/cancel"1 mol"#

#=# #color(blue)"136 g C"# to three sig figs.