Question

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

2.) Using the following unbalanced equation, answer the following questions:

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

a.) Balance the reaction.

b.) How many moles of carbon required to produce 5 moles of carbon dioxide?

c.) How many moles of calcium carbide are produced if 4 moles of carbon react with excess calcium carbonate?

d.) How many moles of carbon dioxide are formed if 55 g of calcium carbonate react with excess carbon?

e.) How many grams of carbon are needed to react with 453 grams of calcium carbonate?

Answer 1

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

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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.